#include<bits/stdc++.h>
using namespace std;
long long c[200006]={0},d,n,j,k,i,l,t=0,ma=0;
struct bian{
	long long x,h;
}b[200006];
bool cmp(bian a,bian b)
{
	return a.x<b.x;
}

int main()
{
	scanf("%lld%lld%lld",&n,&d,&j);
	d=d*2;
	for(i=1;i<=n;i++)
	{
		scanf("%lld%lld",&b[i].x,&b[i].h);
		b[i].h=(b[i].h+j-1)/j;
	}
	sort(b +1,b+n+1 ,cmp);
	queue<long long> q;
	map<long long , long long> m;
	for(i=1;i<=n;i++)
	{
		if(t)
		while(!q.empty())
	    {
	    	if(q.front()<b[i].x)
	    	{
				t-=m[q.front()];
				q.pop();
			}
			else break;
		    
	    }
	    		if(t<b[i].h)
	    		{
					ma+=b[i].h-t;	
					q.push(b[i].x+d);
					m[b[i].x+d]=b[i].h-t;
					t=b[i].h;
					
				}
			
	}
	printf("%lld",ma);

}
### Prompt

Create a solution in CPP for 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;
long long c[200006]={0},d,n,j,k,i,l,t=0,ma=0;
struct bian{
	long long x,h;
}b[200006];
bool cmp(bian a,bian b)
{
	return a.x<b.x;
}

int main()
{
	scanf("%lld%lld%lld",&n,&d,&j);
	d=d*2;
	for(i=1;i<=n;i++)
	{
		scanf("%lld%lld",&b[i].x,&b[i].h);
		b[i].h=(b[i].h+j-1)/j;
	}
	sort(b +1,b+n+1 ,cmp);
	queue<long long> q;
	map<long long , long long> m;
	for(i=1;i<=n;i++)
	{
		if(t)
		while(!q.empty())
	    {
	    	if(q.front()<b[i].x)
	    	{
				t-=m[q.front()];
				q.pop();
			}
			else break;
		    
	    }
	    		if(t<b[i].h)
	    		{
					ma+=b[i].h-t;	
					q.push(b[i].x+d);
					m[b[i].x+d]=b[i].h-t;
					t=b[i].h;
					
				}
			
	}
	printf("%lld",ma);

}
```