#include<bits/stdc++.h>
using namespace std;
int n, d, A;
struct az
{
	int x;
	int h;
}q[200005];

int l[200005];
long long f[200005];
long long sum[200005];

bool cmp(az a, az b)
{
	return a.x < b.x;
}
int main()
{
	scanf("%d%d%d",&n,&d,&A);
	for(int i=1;i<=n;++i)
	    scanf("%d%d",&q[i].x,&q[i].h);
	sort(q+1,q+1+n,cmp);
	int s=1;
    for(int i=1;i<=n;++i)
    {
    	while(s<n && q[s+1].x-q[i].x <= 2*d) ++s;
    	l[i]=s;
    }
	long long ans=0;
	for(int i=1;i<=n;++i)
	{
		sum[i]=sum[i-1]+f[i];
		double p=q[i].h - sum[i] *A;
		//cout<<p<<" "<<i<<endl;
		if (p > 0) 
		{
			f[i] += ceil(p/A);
			f[l[i]+1] -=ceil(p/A);
			sum[i] += ceil(p/A);
			ans+=ceil(p/A);
		}
	}
	cout<<ans<<endl;
	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<bits/stdc++.h>
using namespace std;
int n, d, A;
struct az
{
	int x;
	int h;
}q[200005];

int l[200005];
long long f[200005];
long long sum[200005];

bool cmp(az a, az b)
{
	return a.x < b.x;
}
int main()
{
	scanf("%d%d%d",&n,&d,&A);
	for(int i=1;i<=n;++i)
	    scanf("%d%d",&q[i].x,&q[i].h);
	sort(q+1,q+1+n,cmp);
	int s=1;
    for(int i=1;i<=n;++i)
    {
    	while(s<n && q[s+1].x-q[i].x <= 2*d) ++s;
    	l[i]=s;
    }
	long long ans=0;
	for(int i=1;i<=n;++i)
	{
		sum[i]=sum[i-1]+f[i];
		double p=q[i].h - sum[i] *A;
		//cout<<p<<" "<<i<<endl;
		if (p > 0) 
		{
			f[i] += ceil(p/A);
			f[l[i]+1] -=ceil(p/A);
			sum[i] += ceil(p/A);
			ans+=ceil(p/A);
		}
	}
	cout<<ans<<endl;
	return 0;
}
```