#include<iostream>
#include<algorithm>
#include<queue>
using namespace std;
int N;
long D,A;
vector<pair<int,int> >H;
int main()
{
	cin>>N>>D>>A;
	for(int i=0;i<N;i++)
	{
		int x,h;cin>>x>>h;
		H.push_back(make_pair(x,h));
	}
	sort(H.begin(),H.end());
	queue<pair<long,int> >P;
	long cnt=0,ans=0;
	for(int i=0;i<N;i++)
	{
		while(!P.empty()&&P.front().first<H[i].first)
		{
			cnt-=P.front().second;
			P.pop();
		}
		long nowh=H[i].second;
		nowh-=cnt*A;
		if(nowh>0)
		{
			long t=(nowh+A-1)/A;
			ans+=t;
			cnt+=t;
			P.push(make_pair(H[i].first+D*2,t));
		}
	}
	cout<<ans<<endl;
}

### Prompt

Please formulate 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<iostream>
#include<algorithm>
#include<queue>
using namespace std;
int N;
long D,A;
vector<pair<int,int> >H;
int main()
{
	cin>>N>>D>>A;
	for(int i=0;i<N;i++)
	{
		int x,h;cin>>x>>h;
		H.push_back(make_pair(x,h));
	}
	sort(H.begin(),H.end());
	queue<pair<long,int> >P;
	long cnt=0,ans=0;
	for(int i=0;i<N;i++)
	{
		while(!P.empty()&&P.front().first<H[i].first)
		{
			cnt-=P.front().second;
			P.pop();
		}
		long nowh=H[i].second;
		nowh-=cnt*A;
		if(nowh>0)
		{
			long t=(nowh+A-1)/A;
			ans+=t;
			cnt+=t;
			P.push(make_pair(H[i].first+D*2,t));
		}
	}
	cout<<ans<<endl;
}

```