APPLICATION OF FRONTAL DISPLACEMENT THEORYStep 1

Sw0.250.300.350.400.450.500.550.600.650.700.75

ln a = 6.2717slope b=

Step 3 Plot the water saturation profile as a function of distance andtime

0.25 0.35 0.45 0.55 0.65 0.75

-3

-2

-1

0

1

2

3

4

f(x) = − 11.4738708545726 x + 6.27172510577153

Ln (Kro/Krw) VS Sw

0 50 100 150 200 250 300 350 4000

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

60 days

Linear (60 days)

120 days

Linear (120 days)

240 days

Linear (240 days)

Alternatively:

775779.2

430.9884 days

Q.2 when the water saturation at the producing wellreaches 0.70 (i.e., Sw2 = 0.7):

0 50 100 150 200 250 300 350 4000

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

60 days

Linear (60 days)

120 days

Linear (120 days)

240 days

Linear (240 days)

0.20 0.30 0.40 0.50 0.60 0.70 0.800

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0

0.5

1

1.5

2

2.5

3

Sw

fw

dfw

/dSw

0.20 0.30 0.40 0.50 0.60 0.70 0.800

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0

0.5

1

1.5

2

2.5

3

Sw

fw

dfw

/dSw

APPLICATION OF FRONTAL DISPLACEMENT THEORY

ln(Kro/Krw)=ln a +bSwKro/Krw ln(Kro/Krw) fw (dfw/dsw)

30.23 3.408834809 0.062383 0.67112717 2.833213344 0.105613 1.083825

9.56 2.257587727 0.173266 1.6435945.38 1.682688374 0.271119 2.2674173.02 1.105256831 0.397653 2.748311

1.7 0.530628251 0.539529 2.8505710.96 -0.040821995 0.675276 2.5159980.54 -0.616186139 0.786817 1.9246

0.3 -1.203972804 0.867559 1.3183660.17 -1.771956842 0.920799 0.836782

0.1 -2.302585093 0.953777 0.50585

ln a = 6.2717 a= 529.3766-11.474

(Swf, fwf)= (0.78, 0.59)(dfw/dSw)Swf = 2

This means that the leading edge of the waterfront (stabilized zone) has aconstant saturation of 0.59 and water cut of 78%.

noted that no water saturation with a value less than Swf, i.e.,59%, exists behind the leading edge of the water bank.Assume water saturation values in the range of Swf to (1 – Sor),i.e., 59 to 75%, and calculate the water saturation profile as afunction of time by using

When constructing the water saturation profile, it should be

0.25 0.35 0.45 0.55 0.65 0.75

-3

-2

-1

0

1

2

3

4

f(x) = − 11.4738708545726 x + 6.27172510577153

Ln (Kro/Krw) VS Sw

0.20 0.30 0.40 0.50 0.60 0.70 0.800

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0

0.5

1

1.5

2

2.5

3

Swfw

dfw

/dSw

0 50 100 150 200 250 300 350 4000

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

60 days

Linear (60 days)

120 days

Linear (120 days)

240 days

Linear (240 days)

0.59 2 92.4 184.8 369.60.6 1.96 90.552 181.104 362.208

0.65 1.58 72.996 145.992 291.9840.7 0.85 39.27 78.54 157.08

0.75 0.5 23.1 46.2 92.4

The above example shows that after 240 days of water injection, theleading edge of the water front has moved 365 feet from the injectionwell (235 feet from the producer). The water front (leading edge) willeventually reach the production well and water breakthrough occurs.The example also indicates that at water breakthrough, the leadingedge of the water front would have traveled exactly the entire distancebetween the two wells, i.e., 600 feet. Therefore, to determine the time tobreakthrough, tBT, simply set (x)Swf equal to the distance between theinjector and producer L

t BT = 430.9884431 days

0.5

To Calculate:

a. reservoir water cut in bbl/bbl

b. surface water cut in STB/STBc. reservoir water–oil ratio in bbl/bbld. surface water–oil ratio in STB/STBe. average water saturation in the swept area

0 50 100 150 200 250 300 350 4000

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

60 days

Linear (60 days)

120 days

Linear (120 days)

240 days

Linear (240 days)

f. pore volumes of water injectedg. cumulative water injected in bblAssume that the areal and vertical sweep efficiency are 100%, i.e., EA =1.0 and EV = 1.0.

Solutiona. At Sw= 0.7, fw= 0.922 (reservoir water cut)b. surface water cut,

0.935425

c. reservoir water-oil ratio, 11.82051

d. surface water-oil ratio, 14.48592

(dfw/dSw)Sw = 0.831

f. pore volumes of water injected

1.203369

Averge water saturation in the

swept area

(0.78, 0.59)

This means that the leading edge of the waterfront (stabilized zone) has a

noted that no water saturation with a value less than Swf, i.e.,59%, exists behind the leading edge of the water bank.Assume water saturation values in the range of Swf to (1 – Sor),i.e., 59 to 75%, and calculate the water saturation profile as a

0.20 0.30 0.40 0.50 0.60 0.70 0.800

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0

0.5

1

1.5

2

2.5

3

Sw

fw

dfw

/dSw

387889.6

0.793863

ln(Kro/Krw)=ln a +bSwSw Kro/Krw ln(Kro/Krw) fw (dfw/dsw)

0.25 30.23 3.4088348 0.062383 0.6711274360.30 17 2.8332133 0.105613 1.0838249350.35 9.56 2.2575877 0.173266 1.6435941670.40 5.38 1.6826884 0.271119 2.2674168850.45 3.02 1.1052568 0.397653 2.7483114740.50 1.7 0.5306283 0.539529 2.8505712970.55 0.96 -0.040822 0.675276 2.515998440.60 0.54 -0.6161861 0.786817 1.924600365 Step 1. Plot fw vs. Sw as shown in Figure 14-30 and construct the tangent to the curve. Extrapolate the tangent to fw=1.0 and determine:0.65 0.3 -1.2039728 0.867559 1.3183661680.70 0.17 -1.7719568 0.920799 0.8367822370.75 0.1 -2.3025851 0.953777 0.505850175

ln a = 6.2717 a= 529.3766slope b= -11.474

Step 2. Calculate EDBT

0.634

Step 3. Calculate (Np)BT = Ns x EDBT314624.55

Step 4. Calculate cumulative water injected at breakthrough = PV X 1/(dfw/dSw)swf387889.5

Step 5. Calculate the time to breakthrough: QiBT/iw 430.9883

Step 6. Calculate WORs exactly at breakthrough Bo/Bw(1/fw-1) 4.34492

Step 7. Describe the recovery performance to breakthrough in the followingtabulated form:t, daysWinj = 900t Np = Winj/Bo Qo = iw/Bo WORs Qw = Qo * WO Wp

0 0 0 0 0 0 0100 90000 72000 720 0 0 0200 180000 144000 720 0 0 0300 270000 216000 720 0 0 0400 360000 288000 720 0 0 0431 387891 310312.8 720 4.34 3124.8 0

Step 8. Following the computational procedure as outlined for recovery Recallperformance after breakthrough, construct the followingtable: Sw2 is

Sw2 fw2 dfw/dSw Qi ED Np Winj t,days0.598 0.782942862 1.9499298467 0.512839 0.709315 0.636644197 316061.4 397849.7 442.05520.600 0.786817394 1.92460036532 0.5195884 0.710767 0.638459008 316962.3 403085.8 447.87310.700 0.92079858 0.83678223706 1.195054 0.79465 0.743312471 369016.7 927097.8 1030.1090.800 0.973419754 0.29687525004 3.3684182 0.889533 0.86191673 427897.7 2613148 2903.498

Sw2-

0.20 0.30 0.40 0.50 0.60 0.70 0.800

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0

0.5

1

1.5

2

2.5

3

Sw

fw

dfw

/dSw

t, daysNp Winj Wp WORs Qo Qw0 0 0 0 0 0 0

100 72000 90000 0 0 720 0200 144000 180000 0 0 720 0300 216000 270000 0 0 720 0400 288000 360000 0 0 720 0431 310312.8 387891 0 4.34 720 3124.8442 316061.3749 397849.697676 2718.6069 4.420443 156.2811393 690.8319448 316962.3359 403085.759505 6747.8819 4.523057 153.491476 694.2506

1030 369016.7331 927097.834584 456693.06 14.24759 57.02502212 812.46932903 427897.6986 2613148.11491 2037525.5 44.87981 19.13777693 858.8998

0 500 1000 1500 2000 2500 3000 35000

500

1000

1500

2000

2500

3000

3500

0

500000

1000000

1500000

2000000

2500000

3000000

Water Flood Performance Curve

WOR Qo Qw Np Winj Wp

Time, days

Step 1. Plot fw vs. Sw as shown in Figure 14-30 and construct the tangent to the curve. Extrapolate the tangent to fw=1.0 and determine:Swf = SwBT = 0.59fwf = fwBT = 0.780(dfw/dSw)swf = 2QiBT = 1/2 0.5

0.707

a = 529.3766b = -11.474

Wp WORs Qo Qw2718.607 4.420443 156.2811 690.83196747.882 4.523057 153.4915 694.2506456693.1 14.24759 57.02502 812.46932037525 44.87981 19.13778 858.8998

0.20 0.30 0.40 0.50 0.60 0.70 0.800

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0

0.5

1

1.5

2

2.5

3

Sw

fw

dfw

/dSw

0 500 1000 1500 2000 2500 3000 35000

500

1000

1500

2000

2500

3000

3500

0

500000

1000000

1500000

2000000

2500000

3000000

Water Flood Performance Curve

WOR Qo Qw Np Winj Wp

Time, days

Step 1. Plot fw vs. Sw as shown in Figure 14-30 and construct the tangent to the curve. Extrapolate the tangent to fw=1.0 and determine:

ln(Kro/Krw)=ln a +bSwSw Kro/Krw ln(Kro/Krw)

0.1 #DIV/0! #DIV/0!0.3 5.328571 1.6730831780.4 1.242604 0.217208816

0.45 0.654867 -0.423322730.5 0.333333 -1.09861229

0.55 0.162234 -1.818715280.6 0.069328 -2.66891029

0.65 0.02 -3.912023010.7 0 Err:502

0.2 0.3 0.4 0.5 0.6 0.7

-5

-4

-3

-2

-1

0

1

2

3

f(x) = − 15.34455074 x + 6.415344067

0.4 0.45 0.5 0.55 0.6 0.65 0.70

0.5

1

1.5

2

2.5

Swf

dfw

/dSw

Note: because Sgi=0, the cumulative water injected at breakthrough will displace an equivalent volume of oil

25.49.7282 11.5282

0.648 150,816 232740.7 1.80.663 154,307 232740.6

Step 1. Calculate the distance between the producer and the injector:Total Area = 40 acre 1 acre = 43560

1 Qtr of area 435600Qtr side lengt 660

d = 933.381

step 2 Calculate the initial (base) injection rate

269.0165

Step 3 Notice that the cumulative water injected of 60,000 bbl is lessthan the amount of cumulative water injected at breakthrough of103,020 bbl; therefore, M = 0.8 (remains constant until breakthrough)and EA

0.4176Step 4. Calculate the conductance ratios from Figure ( using M and EA) 0.92

M = 0.8

Step 5. Calculate the water-injection rate when the cumulative water

injected reaches 60,000 bbliw= gamma x ibase

247.4951624Step 6. After breakthrough when the cumulative water injected reaches144,230 barrels of water, the average water saturation in theswept area is 59%

0.59Step 7. Determine the water relative permeability krw at 0.59 water saturation to give krw = 0.45.

Step 8. Calculate the mobility ratio after breakthrough when Winj =144,230

M = 0.9

Step 9. Calculate the areal sweep efficiency when Winj = 144,230 EA = 0.845.

0.809523Step 10. Determine the conductance ratio from Figure 14-42: γ = 0.96

Step 11. Calculate the water injection rate iw = (269.1) (0.96) = 258.3 bbl/day

The conductance ratio can be expressed more conveniently in amathematical form. For an areal sweep efficiency of100%, i.e., EA = 1.0: CONDUCTANCE RATIO = M

Step 1 Calculate stock tank oil in place at start of flood, Ns:Ns = PV Soi/Boi

Step 2 Calculate injected water at interference WiiWii =

Step 3 Simplify the calculations by expressing outer radii of the oil and water banksro =

3.452 (Winj)^(1/2)

r =0.562ro

Step 4 Express the injectivityiw =

Step 5. Perform the required calculation for “stage one” in the followingtabulated form:Winj ro r

500 77.2 43.45000 244.1 137.2

S-w2

¶hØSgirei^2/5.615

(5.615Winj/(¶hØSgi))^(1/2)

ro (Sgi/(S-Wbt - Swi))

10000 345.2 194.015000 422.8 237.620000 488.2 274.425000 545.8 306.730000 597.9 336.035000 645.8 362.936572 660.2 371.0

Step 1. Calculate the cumulative water injected at fill-up from Equation14-81:Wif = (PV)Sgi = 310,320(0.15) Step 2. Calculate the areal sweep efficiency at fill-up

EA 0.323988002Step 3. Given a mobility ratio M of 0.8 (Example 14-11) and EA of0.324, calculate the conductance ratio at the fill-up from Figure

Note that the conductance ratio can only be determinedwhen the flood pattern is completely filled with liquids, whichoccurs at the fill-up stage.

γ = 0.96Step 4. Calculate the initial (base) injection rate

ln(Kro/Krw)=ln a +bSwfw (dfw/dsw)0.014955 0.22605007110.246261 2.84828743080.602489 3.675065945460.765504 2.754542733180.875484 1.672779920330.938057 0.891636762940.970254 0.442881440340.985965 0.212338106550.993435 0.10008538183

b= -15.345a= 611.124070541

Phase 1. Initial CalculationsStep 1 Calculate Pore Volume and oil volume at start of flood

PV = 310320

Ns = 232740Step 2 Plot fw vs Sw and determine:

Swf = SwBT 0.469 QBT 0.462962962962963

fwf =fwBT 0.798 0.563(dfw/dSw)swf 2.16

Step 3 Determine Kro and Krw at Swi and S-wBT from the relativity permeability dataKro @ 0.1 1

0.4Step 4 Calculate the mobility ratio M

0.8Step 5 Calculate the areal sweep efficiency at breakthrough

: from graph using M = 0.8EABT = 0.71702

Step 6 Select several values of Sw2 between Swf and (1 - Sor) and calculate dfw/dSwSw2 fw2 dfw/dSw

0.469 0.798 2.3255729290.495 0.8668768378958 1.770834265

0.52 0.9052738314133 1.3158816510.546 0.9343931506571 0.9406882540.572 0.955005944354 0.6593683680.622 0.9785938238192 0.321446320.649 0.9857514933172 0.2155279940.674 0.9902469778222 0.148200487

0.7 0.9934345503244 0.100085377

Phase 2. Calculation of Recovery Performance to Breakthrough

Step 1

7758AhØ

S-wBT

Krw @ 0.563

QiBT = (S-Wbt - Swi)

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.80

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0

0.5

1

1.5

2

2.5

3

3.5

4

0.4 0.45 0.5 0.55 0.6 0.65 0.70

0.5

1

1.5

2

2.5

Swf

dfw

/dSw

0.4631/(dfw/dSw)swf 0.463

Step 2 Calculate cumulative water injected at breakthroughWiBT = (PV) QiBT EABT103020.114283

step 3 Calculate Time to BreakthroughTbt = WiBT/ iw

382.97440254 daysStep 4 Calculate the displacement efficiency at breakthrough

0.51444444444Step 5 Calculate the cumulative oil production at breakthrough

(Np)BT = Ns EDBT EABT85850.095236 STB

Note: because Sgi=0, the cumulative water injected at breakthrough will displace an equivalent volume of oil(Np)BT = WiBT/ Bo103,020/1.2

85850Step 6 Calculate the surface water cut WOR, exactly at breakthrough

1.11151187122Λ = 0.2749(WiBT/Winj)

0.27490.305554613

WORs=

1.4915933240806 stb/stbStep 7 Set up the following table to describe the oil recovery performance to breakthrough (NB: Sgi = 0)

Winj t =Winj/iw Np= Winj/Bo Qo = iw/Bo WORsbbl days stb stb/d stb/stb

0 0 0 0 020000 74 16667 224 040000 149 33333 224 060000 223 50000 224 080000 297 66667 224 0

103020 383 85850 224 1.49

Phase 3. Oil Recovery Calculations After Breakthrough

Winj t =Winj/iw Winj/WiBT Qi/QiBT

103,020 383 1.0 0.717 1 120,000 446 1.2 0.759 1.193 140,000 520 1.4 0.801 1.375 160,000 595 1.6 0.838 1.548

EDBT = (S-Wbt - Swi) / (1-Swi)

E= (Swf- Swi)/ EABT (S-Wbt-Swi)

(ΔNp)newly = EΛ

EA

.=EABT + 0.633 log (Winj/WiBT)

Table Qi/QiBT as a function of Winj/WiBT and EABT

180,000 669 1.7 0.870 1.65 200,000 743 1.9 0.899 1.8 250,000 929 2.4 0.961 2.183 300,000 1115 2.9 1.000 2.6 350,000 1301 3.4 1.000 2.9 400,000 1487 3.9 1.000 3.336 500,000 1859 4.9 1.000 4 600,000 2230 5.8 1.000 4.59 800,000 2974 7.8 1.000 6 1,000,000 3717 9.7 1.000 7.3 1,500,000 5576 14.6 1.000 11

before breakthrough:

Calculate stock tank oil in place at start of flood, Ns:193950 stb

Calculate injected water at interference Wii36572 bbl

Simplify the calculations by expressing outer radii of the oil and water banks

3.452 (Winj)^(1/2)

3340/(1.25lnr+ln(ro/r))Step 5. Perform the required calculation for “stage one” in the following

iw (iw)avg Δt= (Winj/iw)avg t = ∑Δt631.5 0.8 0.8496.4 564.0 8.0 8.8

(5.615Winj/(¶hØSgi))^(1/2)

ro (Sgi/(S-Wbt - Swi))

466.4 481.4 10.4 19.2450.5 458.4 10.9 30.1439.8 445.1 11.2 41.3431.9 435.8 11.5 52.8425.6 428.7 11.7 64.4420.4 423.0 11.8 76.2419.0 419.7 3.7 80.0

Step 1. Calculate the cumulative water injected at fill-up from Equation

46550 bblStep 2. Calculate the areal sweep efficiency at fill-up

Step 3. Given a mobility ratio M of 0.8 (Example 14-11) and EA of0.324, calculate the conductance ratio at the fill-up from Figure Np is negligible or zero prior to fill up

Note that the conductance ratio can only be determinedwhen the flood pattern is completely filled with liquids, whichoccurs at the fill-up stage.

Step 4. Calculate the initial (base) injection rate

0.46296296296

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.80

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0

0.5

1

1.5

2

2.5

3

3.5

4

Set up the following table to describe the oil recovery performance to breakthrough (NB: Sgi = 0)Qw=Qo* WORs Wpstb/day stb

0 00 00 00 00 0

334 0

Qi (dfw/dSw)Sw2 Sw2 fw2 ED Np

(Qi/QiBT) QiBT 1/Qi Assumed Ns ED EA EV

0.463 2.160 0.48 0.838003 0.55500440253 0.50556045 84,368 0.552 1.810 0.49 0.857771 0.56856157258 0.52062397 91,963 0.637 1.571 0.5 0.875484 0.57926968439 0.53252187 99,317 0.717 1.395 0.515 0.898487 0.5877570118 0.54195224 105,706

S-w2

from plots dfw/dSw vs

Sw

1/((1+((µw/

µo)aexp(bSw))))

Sw2+((1-fw2)/(dfw/dSw)Sw2)

(S-w - Swi)/ (1-Swi)

0.764 1.309 0.52 0.905274 0.59236605649 0.5470734 110,828 0.833 1.200 0.525 0.911652 0.59862939748 0.55403266 115,973 1.011 0.989 0.54 0.928518 0.61224872351 0.56916525 127,266 1.204 0.831 0.559 0.945613 0.62447144572 0.58274605 135,628 1.343 0.745 0.56 0.946396 0.63197342417 0.59108158 137,568 1.545 0.647 0.57 0.953669 0.64156193874 0.60173549 140,048 1.852 0.540 0.59 0.96549 0.65391186756 0.61545763 143,242 2.125 0.471 0.6 0.970254 0.66321629083 0.62579588 145,648 2.778 0.360 0.615 0.976224 0.68104929077 0.64561032 150,259 3.380 0.296 0.62 0.977941 0.69455586366 0.66061763 153,752 5.093 0.196 0.7 0.993435 0.7334378352 0.70381982 163,807

Wp WORs Qo Qw

(Winj - NpBo)/Bw Qo WORs

0 1.6704537685755 93.71341 156.54399644 1.7679093527729 90.63619 160.2366

20819 1.8610383897936 87.87868 163.545633152 1.991064598031 84.29789 167.8425

fw2(1-Δnpnew)/((1-(fw2(1-∆Npnew)))

(Bo/Bw)

iw/(Bo + (Bw

WORs))

47006 2.0315679915708 83.24132 169.110460833 2.0705780337115 82.24846 170.301997280 2.178427684958 79.62284 173.4526

137246 2.2952444074004 76.96171 176.6459184918 2.3007946348469 76.8397 176.7924231942 2.3531424356355 75.70763 178.1508328110 2.4416641484972 73.86733 180.3592425223 2.4785904320077 73.12584 181.249619689 2.5259485685746 72.19638 182.3643815497 2.5397961260703 71.92905 182.6851

1303432 2.6695452566461 69.51721 185.5793

qt336336304309272248208197184176184

2754

method of least squares

-0.0004144724

Time, months MMScf/month0 12401 11932 11483 11044 10665 10236 9867 9498 9119 880

10 843

11 813

0 100000 200000 300000 400000 500000 600000 700000 8000000

50100150200250300350400

Gp vs qt Linear (Gp vs qt)

12 782

Gp qt Gp Gp^216000 5376000 25600000032000 10752000 102400000048000 14592000 230400000096000 29664000 9216000000

160000 43520000 25600000000240000 59520000 57600000000304000 63232000 92416000000352000 69344000 123904000000368000 67712000 135424000000384000 67584000 147456000000400000 73600000 160000000000

2400000 5.05E+08 755200000000

A straight line implies exponential decline curve

when gt=80, Gp =633,600

qi = 344

D 0.00042Diy 0.1512

t(ln(qi/qt)) t^2 Step 1. A plot of qt versus t on a semi-log scale0 0 indicates an exponential decline.

0.0386402365 10.1541801634 40.3485142953 90.6047922155 160.9618594632 25

1.375261824 36

0 100000 200000 300000 400000 500000 600000 700000 8000000

50100150200250300350400

Gp vs qt Linear (Gp vs qt)

0 1 2 3 4 5 6 7100

1000

10000

3.4832481979 91calculate the time, ta, to reach aneconomic flow rate, qa, of 30 MMscf/month, and the correserves, Gpa:

ta 97.4215175857351 8.11846Gpa 11.9070204277936

0 1 2 3 4 5 6 7100

1000

10000

A straight line implies exponential decline curve

Step 1. A plot of qt versus t on a semi-log scale

Step 2. Determine the initial decline rate, Di, by selecting a point on thestraight line and substituting the coordinates of the point inD = Ln(qi/qt)/t

0.03820171733Alternatively0.038277452725

Use Equations 16-3 and 16-7 to calculate qt and Gp(t), and tabulate theresults as follows:

Time, months Gp(t)Actual qt, MMScf/month

Calculated qt, MMScf/month

0 1 2 3 4 5 6 7100

1000

10000

0 1240 1240 12401 1193 1194 12172 1148 1149 2388

economic flow rate, qa, of 30 MMscf/month, and the cor 3 1104 1106 35154 1066 1064 46005 1023 1024 56446 986 986 66497 949 949 76168 911 913 85479 880 879 9444

10 843 846 1030611 813 815 1113712 782 784 11936

0 1 2 3 4 5 6 7100

1000

10000