#include<bits/stdc++.h>
using namespace std;
long long n,d,g,k,a[1000005],h[1000005],sum,j,ans,i;
pair<long long,long long>x[200005],p[200005];
int main(){
	
	cin>>n>>d>>g;
	for(k=1;k<=n;k++){
		cin>>a[k]>>h[k];
		p[k].first=a[k];
		p[k].second=h[k];
	}
	sort(p+1,p+n+1);
	for(k=1;k<=n;k++){
		a[k]=p[k].first;
		h[k]=p[k].second;
		while(x[j].first<a[k] && j<=i) {
		sum-=x[j].second;	j++; 
		}
		
		h[k]-=sum; 
		if(h[k]>0){
			i++;
			x[i].first=a[k]+2*d;
			x[i].second=h[k]/g;
			if(h[k]%g>0) x[i].second++;
			
			ans+=x[i].second;
		           x[i].second*=g;
			sum+=x[i].second; 
			
		}
	
		
		
	}
	cout<<ans;
}
### 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;
long long n,d,g,k,a[1000005],h[1000005],sum,j,ans,i;
pair<long long,long long>x[200005],p[200005];
int main(){
	
	cin>>n>>d>>g;
	for(k=1;k<=n;k++){
		cin>>a[k]>>h[k];
		p[k].first=a[k];
		p[k].second=h[k];
	}
	sort(p+1,p+n+1);
	for(k=1;k<=n;k++){
		a[k]=p[k].first;
		h[k]=p[k].second;
		while(x[j].first<a[k] && j<=i) {
		sum-=x[j].second;	j++; 
		}
		
		h[k]-=sum; 
		if(h[k]>0){
			i++;
			x[i].first=a[k]+2*d;
			x[i].second=h[k]/g;
			if(h[k]%g>0) x[i].second++;
			
			ans+=x[i].second;
		           x[i].second*=g;
			sum+=x[i].second; 
			
		}
	
		
		
	}
	cout<<ans;
}
```