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: Accepted)

#include<stdio.h>
#include<stdlib.h>
#define ll long long

typedef struct monster{
	int x,h;
}M;

M m[200010];
ll cum[200010];

int c(const void*p,const void*q){
	if((*(M*)p).x<(*(M*)q).x)return -1;
	return 1;
}


int main(){
	int n,d,a;
	scanf("%d%d%d",&n,&d,&a);
	for(int i=0;i<n;i++)scanf("%d%d",&(m[i].x),&(m[i].h));
	qsort(m,n,sizeof(M),c);

	ll ans=0;
	int r=0;
	for(int i=0;i<n;i++){
		if(i)cum[i]+=cum[i-1];
		if(m[i].h-cum[i]<=0)continue;
		while(r<n&&m[r].x-m[i].x<=2*d)r++;
		int cnt=(m[i].h-cum[i]+(a-1))/a;
		ans+=cnt;
		cum[i]+=a*cnt;
		cum[r]-=a*cnt;
	}
	printf("%lld\n",ans);
}