#include <bits/stdc++.h>
using namespace std;

int main(){
	int N,D,A;
	cin >> N >> D >> A;
	vector<pair<long long,long long>> XH;
	for(int i=0;i<N;i++){
		long long x,h;
		cin >> x >> h;
		XH.push_back(make_pair(x,(h+A-1)/A));
	}
	sort(XH.begin(),XH.end());
	
	queue<pair<long long,long long>> que;
	long long ans=0,d=0;
	
	for(int i=0;i<N;i++){
		long long X=XH[i].first,H=XH[i].second;
		
		while(!que.empty()){
			if(que.front().first>=X)break;
			d-=que.front().second;
			que.pop();
		}
		
		H=max(0LL,H-d);
		ans+=H;
		d+=H;
		if(H>0)que.push(make_pair(X+D*2,H));
	}
	
	cout << ans << endl;
	
	return 0;
}
### Prompt

Please create a solution in CPP 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 main(){
	int N,D,A;
	cin >> N >> D >> A;
	vector<pair<long long,long long>> XH;
	for(int i=0;i<N;i++){
		long long x,h;
		cin >> x >> h;
		XH.push_back(make_pair(x,(h+A-1)/A));
	}
	sort(XH.begin(),XH.end());
	
	queue<pair<long long,long long>> que;
	long long ans=0,d=0;
	
	for(int i=0;i<N;i++){
		long long X=XH[i].first,H=XH[i].second;
		
		while(!que.empty()){
			if(que.front().first>=X)break;
			d-=que.front().second;
			que.pop();
		}
		
		H=max(0LL,H-d);
		ans+=H;
		d+=H;
		if(H>0)que.push(make_pair(X+D*2,H));
	}
	
	cout << ans << endl;
	
	return 0;
}
```