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

### Prompt

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

```