#include<bits/stdc++.h>
using namespace std;
int n,d,a;
long long ans=0;
struct node
{
	int x,h,hh;
}m[200005];
bool cmp(node u,node v)
{
	return u.x<v.x;
}
int main()
{
	cin>>n>>d>>a;
	for(int i=0;i<n;i++)
	{
		cin>>m[i].x>>m[i].h;
		m[i].h=(m[i].h+a-1)/a;
	}
	sort(m,m+n,cmp);
	int l=0,cnt=0;
	for(int i=0;i<n;i++)
	{
		while(m[i].x-m[l].x>d*2)
		{
			cnt-=m[l].h;
			l++;
		}
		ans+=max(0,m[i].h-cnt);
		m[i].h-=cnt;
		m[i].h=max(0,m[i].h);
		cnt+=m[i].h;
	}
	cout<<ans<<endl;
	return 0;
}

### Prompt

Your task is to create 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;
int n,d,a;
long long ans=0;
struct node
{
	int x,h,hh;
}m[200005];
bool cmp(node u,node v)
{
	return u.x<v.x;
}
int main()
{
	cin>>n>>d>>a;
	for(int i=0;i<n;i++)
	{
		cin>>m[i].x>>m[i].h;
		m[i].h=(m[i].h+a-1)/a;
	}
	sort(m,m+n,cmp);
	int l=0,cnt=0;
	for(int i=0;i<n;i++)
	{
		while(m[i].x-m[l].x>d*2)
		{
			cnt-=m[l].h;
			l++;
		}
		ans+=max(0,m[i].h-cnt);
		m[i].h-=cnt;
		m[i].h=max(0,m[i].h);
		cnt+=m[i].h;
	}
	cout<<ans<<endl;
	return 0;
}

```