#include<iostream>
#include<algorithm>
#include<cstdio>
using namespace std;
struct monster
{
	long long x,h;
}a[500005];
bool cmp(monster a,monster b)
{
	return a.x<b.x;
}
int main()
{
	long long n,d,hurt,ans=0,l=0,r=0,now=0;
	scanf("%lld%lld%lld",&n,&d,&hurt);
	for(long long i=0;i<n;i++) scanf("%lld%lld",&a[i].x,&a[i].h);
	sort(a,a+n,cmp);
	while(r<n)
	{
		ans+=(a[r].h+hurt-1)/hurt;
		now+=(a[r].h+hurt-1)/hurt;
		r++;
		while(a[r].x>2*d+a[l].x&&l<r)
		{
			now-=(a[l].h+hurt-1)/hurt;
			l++;
		}
		a[r].h-=now*hurt;
		if(a[r].h<0) a[r].h=0;
	}
	printf("%lld",ans);
	return 0;
}

### Prompt

Generate a Cpp solution to the following problem:

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.

Examples

Input

3 3 2
1 2
5 4
9 2


Output

2


Input

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


Output

5


Input

3 0 1
300000000 1000000000
100000000 1000000000
200000000 1000000000


Output

3000000000

### Response

```cpp
#include<iostream>
#include<algorithm>
#include<cstdio>
using namespace std;
struct monster
{
	long long x,h;
}a[500005];
bool cmp(monster a,monster b)
{
	return a.x<b.x;
}
int main()
{
	long long n,d,hurt,ans=0,l=0,r=0,now=0;
	scanf("%lld%lld%lld",&n,&d,&hurt);
	for(long long i=0;i<n;i++) scanf("%lld%lld",&a[i].x,&a[i].h);
	sort(a,a+n,cmp);
	while(r<n)
	{
		ans+=(a[r].h+hurt-1)/hurt;
		now+=(a[r].h+hurt-1)/hurt;
		r++;
		while(a[r].x>2*d+a[l].x&&l<r)
		{
			now-=(a[l].h+hurt-1)/hurt;
			l++;
		}
		a[r].h-=now*hurt;
		if(a[r].h<0) a[r].h=0;
	}
	printf("%lld",ans);
	return 0;
}

```