A sum of *p* rubles is charged from Arkady's mobile phone account every day in the morning. Among the following *m* days, there are *n* days when Arkady will top up the account: in the day *d*<sub class="lower-index">*i*</sub> he will deposit *t*<sub class="lower-index">*i*</sub> rubles on his mobile phone account. Arkady will always top up the account before the daily payment will be done. There will be no other payments nor tops up in the following *m* days.

Determine the number of days starting from the 1-st to the *m*-th such that the account will have a negative amount on it after the daily payment (i. e. in evening). Initially the account's balance is zero rubles.
The first line contains three integers *n*, *p* and *m* (1 ≤ *n* ≤ 100 000, 1 ≤ *p* ≤ 10<sup class="upper-index">9</sup>, 1 ≤ *m* ≤ 10<sup class="upper-index">9</sup>, *n* ≤ *m*) — the number of days Arkady will top up the account, the amount of the daily payment, and the number of days you should check.

The *i*-th of the following *n* lines contains two integers *d*<sub class="lower-index">*i*</sub> and *t*<sub class="lower-index">*i*</sub> (1 ≤ *d*<sub class="lower-index">*i*</sub> ≤ *m*, 1 ≤ *t*<sub class="lower-index">*i*</sub> ≤ 10<sup class="upper-index">9</sup>) — the index of the day when Arkady will make the *i*-th top up, and the amount he will deposit on this day. It is guaranteed that the indices of the days are distinct and are given in increasing order, i. e. *d*<sub class="lower-index">*i*</sub> &gt; *d*<sub class="lower-index">*i* - 1</sub> for all *i* from 2 to *n*.
In the first example the balance will change as following (remember, initially the balance is zero):
 1.  in the first day 6 rubles will be charged, the balance in the evening will be equal to  - 6; 1.  in the second day Arkady will deposit 13 rubles, then 6 rubles will be charged, the balance in the evening will be equal to 1; 1.  in the third day 6 rubles will be charged, the balance in the evening will be equal to  - 5; 1.  in the fourth day Arkady will deposit 20 rubles, then 6 rubles will be charged, the balance in the evening will be equal to 9; 1.  in the fifth day 6 rubles will be charged, the balance in the evening will be equal to 3; 1.  in the sixth day 6 rubles will be charged, the balance in the evening will be equal to  - 3; 1.  in the seventh day Arkady will deposit 9 rubles, then 6 rubles will be charged, the balance in the evening will be equal to 0. 
Thus, in the end of the first, third and sixth days the balance will be negative in the end of the day.
3 6 7
2 13
4 20
7 9
3
5 4 100
10 70
15 76
21 12
30 100
67 85
26
14 25 100
1 209
2 224
3 58
4 31
5 135
6 16
7 130
8 113
9 230
10 60
11 209
12 185
13 118
14 16
31
1 1 1
1 1
0
1 2 1
1 1
1
1 2 1
1 3
0
1 1000000000 1
1 1000000000
0
1 100000000 1
1 99999999
1
10 1 10
1 1000000000
2 1000000000
3 1000000000
4 1000000000
5 1000000000
6 1000000000
7 1000000000
8 1000000000
9 1000000000
10 1000000000
0
8 518 10
1 1
2 650
3 436
4 525
6 1163
7 416
8 512
9 500
10
14 4115 100
18 85238
20 1
28 38150
32 16316
33 269
42 47018
48 18222
53 22346
57 13571
58 1717
65 40035
90 93955
93 10747
96 15415
70
8 518 10
1 1
2 581
3 638
4 420
6 1447
7 31
8 574
9 689
10
14 4115 100
18 96022
20 1
28 28266
32 18061
33 2
42 46869
48 20282
53 22181
57 14156
58 1415
65 29530
90 122045
93 5
96 23547
59
8 518 10
1 1
2 528
3 668
4 459
6 1103
7 413
8 549
9 550
10
14 4115 100
18 72959
20 5578
28 36578
32 12797
33 4242
42 42918
48 21810
53 19442
57 14274
58 5036
65 34539
90 98082
93 12170
96 13730
98
8 518 10
1 1
2 820
3 378
4 513
6 915
7 692
8 398
9 538
10
14 4115 100
18 136900
20 1
28 2
32 79818
33 2
42 8712
48 21894
53 28023
57 14205
58 5
65 40631
90 154277
93 5
96 3
26
8 518 10
1 1
2 524
3 676
4 389
6 1013
7 623
8 529
9 606
10
14 4115 100
18 103373
20 1
28 20482
32 14482
33 2321
42 43952
48 19687
53 24828
57 9584
58 6392
65 28693
90 102250
93 15860
96 11260
66
n, p, m = map(int, input().split())
prev=1
ans=0
sum=0
for _ in range(n):
    a,T=map(int,input().split())
    can=sum//p
    day=a-prev
    if(day>0):
        if(day>can):
            ans=ans+day-can
    prev=a
    sum=sum+T
    sum=sum-(day*p)
can=sum//p
day=a-prev
if(day>0):
    if(day>can):
       ans=ans+day-can
print(ans)

