#include<bits/stdc++.h>
using namespace std;
#define pii pair<int,int>
#define LL long long
const int N=2e5+5;

int n,d,a;
pii p[N];
int r[N];
LL ans;

int main()
{
	scanf("%d%d%d",&n,&d,&a);
	for(int i=1;i<=n;i++)
		scanf("%d%d",&p[i].first,&p[i].second);
	sort(p+1,p+n+1);
	int now=1;
	for(int i=1;i<=n;i++)
	{
		p[i].second-=r[i];
		while(now<=n && p[now].first-2*d<=p[i].first) now++;
		if(p[i].second>0)
		{
			int num=(p[i].second+a-1)/a;
			ans+=num;
			r[i]+=num*a;
			r[now]-=num*a;
		}
		r[i+1]+=r[i];
	}
	printf("%lld\n",ans);
	
	return 0;
}
### Prompt

Please provide a Cpp coded solution to the problem described below:

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;
#define pii pair<int,int>
#define LL long long
const int N=2e5+5;

int n,d,a;
pii p[N];
int r[N];
LL ans;

int main()
{
	scanf("%d%d%d",&n,&d,&a);
	for(int i=1;i<=n;i++)
		scanf("%d%d",&p[i].first,&p[i].second);
	sort(p+1,p+n+1);
	int now=1;
	for(int i=1;i<=n;i++)
	{
		p[i].second-=r[i];
		while(now<=n && p[now].first-2*d<=p[i].first) now++;
		if(p[i].second>0)
		{
			int num=(p[i].second+a-1)/a;
			ans+=num;
			r[i]+=num*a;
			r[now]-=num*a;
		}
		r[i+1]+=r[i];
	}
	printf("%lld\n",ans);
	
	return 0;
}
```