Problem:
Score : 600 points

Problem Statement

Silver Fox is fighting with N monsters.

The monsters are standing in a row, and we can assume them to be standing on a number line. The i-th monster, standing at the coordinate X_i, has the health of H_i.

Silver Fox can use bombs to attack the monsters.
Using a bomb at the coordinate x decreases the healths of all monsters between the coordinates x-D and x+D (inclusive) by A.
There is no way other than bombs to decrease the monster's health.

Silver Fox wins when all the monsters' healths become 0 or below.

Find the minimum number of bombs needed to win.

Constraints

1 \leq N \leq 2 \times 10^5

0 \leq D \leq 10^9

1 \leq A \leq 10^9

0 \leq X_i \leq 10^9

1 \leq H_i \leq 10^9

X_i are distinct.

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N D A
X_1 H_1
:
X_N H_N

Output

Print the minimum number of bombs needed to win.

Sample Input 1

3 3 2
1 2
5 4
9 2

Sample Output 1

2

First, let us use a bomb at the coordinate 4 to decrease the first and second monsters' health by 2.

Then, use a bomb at the coordinate 6 to decrease the second and third monsters' health by 2.

Now, all the monsters' healths are 0.
We cannot make all the monsters' health drop to 0 or below with just one bomb.

Sample Input 2

9 4 1
1 5
2 4
3 3
4 2
5 1
6 2
7 3
8 4
9 5

Sample Output 2

5

We should use five bombs at the coordinate 5.

Sample Input 3

3 0 1
300000000 1000000000
100000000 1000000000
200000000 1000000000

Sample Output 3

3000000000

Watch out for overflow.

Solution in C (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#include <stdio.h>
#include <stdlib.h>

struct xhtyp {long x; long h;};

int asbc(const void *xx, const void *y) {
  return (*(struct xhtyp *)xx).x - (*(struct xhtyp *)y).x;
}

int main(){
   int i,j,n;
   long d,a,s,m;
   struct xhtyp xh[200000];
    
   scanf("%d %ld %ld",&n,&d,&a);
   for (i=0;i<n;i++) scanf("%ld %ld",&(xh[i].x),&(xh[i].h));
   qsort(xh,n,sizeof(struct xhtyp),asbc);
   s=0;
   for (i=0;i<n;i++) {
      if (xh[i].h<=0) continue;
      m=(xh[i].h+a-1)/a;
      s+=m;
      for (j=i+1;j<n;j++)
         if (xh[j].x<=xh[i].x+d+d) xh[j].h-=m*a;
         else                      break;
   }
   printf("%ld\n",s);
   return 0;
}