#include <bits/stdc++.h>
#define int long long
#define F first
#define S second
using namespace std;
typedef pair<int,int>pii;
int n,d,a,dd;
pii x[400010];
int ans,tp;
int cha[400010];
signed main(){
	cin>>n>>d>>a;
	dd=d+d;
	for(int i=1;i<=n;i++)cin>>x[i].F>>x[i].S;
	sort(x+1,x+n+1);
	for(int l=1,r=1,cur=0;l<=n;l++){
		while(r<n&&x[r+1].F-x[l].F<=dd)r++;
		int need=(x[l].S-cha[l]*a+a-1)/a;
		if(need<0)need=0;
		ans+=need;
		
		cha[l]+=need,cha[r+1]-=need;//
		cha[l+1]+=cha[l];
		
				
//		cout<<cur<<endl;
//		for(int i=1;i<=n+1;i++)cout<<cha[i]<<' ';
//		cout<<endl;
	}
	cout<<ans<<endl;
	return 0;
}
/*
200010 1 1
1 
*/
### 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>
#define int long long
#define F first
#define S second
using namespace std;
typedef pair<int,int>pii;
int n,d,a,dd;
pii x[400010];
int ans,tp;
int cha[400010];
signed main(){
	cin>>n>>d>>a;
	dd=d+d;
	for(int i=1;i<=n;i++)cin>>x[i].F>>x[i].S;
	sort(x+1,x+n+1);
	for(int l=1,r=1,cur=0;l<=n;l++){
		while(r<n&&x[r+1].F-x[l].F<=dd)r++;
		int need=(x[l].S-cha[l]*a+a-1)/a;
		if(need<0)need=0;
		ans+=need;
		
		cha[l]+=need,cha[r+1]-=need;//
		cha[l+1]+=cha[l];
		
				
//		cout<<cur<<endl;
//		for(int i=1;i<=n+1;i++)cout<<cha[i]<<' ';
//		cout<<endl;
	}
	cout<<ans<<endl;
	return 0;
}
/*
200010 1 1
1 
*/
```