#include <bits/stdc++.h>
using namespace std;
long long n,d,b,j,k[200005],k2[200005],w,sum,ans,sum1[200005],ii;
pair <long long, long long> a[200005];
int main()
{
	cin>>n>>d>>b;
	for (int i=1; i<=n; i++)
	{
		cin>>a[i].first>>a[i].second;
	}
	sort(a+1, a+n+1);
	j=1;
	for (int i=1; i<=n;i++)
	{
		
			while(k[j]<a[i].first  && j<=ii)
		{
			sum-=sum1[j];
			j++;
		} 
		
		a[i].second-=sum;  
		if(a[i].second>0){
		ii++;
		k[ii]=a[i].first+2*d; 
		if (a[i].second%b==0)
		{
			w=a[i].second/b;
		}
		else
		w=a[i].second/b+1;
		sum+=w*b;
		sum1[ii]=w*b;
	ans+=w; 
	} }
	cout<<ans<<endl;
}
### Prompt

Construct a CPP code solution to the problem outlined:

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,b,j,k[200005],k2[200005],w,sum,ans,sum1[200005],ii;
pair <long long, long long> a[200005];
int main()
{
	cin>>n>>d>>b;
	for (int i=1; i<=n; i++)
	{
		cin>>a[i].first>>a[i].second;
	}
	sort(a+1, a+n+1);
	j=1;
	for (int i=1; i<=n;i++)
	{
		
			while(k[j]<a[i].first  && j<=ii)
		{
			sum-=sum1[j];
			j++;
		} 
		
		a[i].second-=sum;  
		if(a[i].second>0){
		ii++;
		k[ii]=a[i].first+2*d; 
		if (a[i].second%b==0)
		{
			w=a[i].second/b;
		}
		else
		w=a[i].second/b+1;
		sum+=w*b;
		sum1[ii]=w*b;
	ans+=w; 
	} }
	cout<<ans<<endl;
}
```