#include<iostream>
#include<string>
#include<vector>
#include<iomanip>
#include<algorithm>
#include<queue>
#include<stack>
#include<map>
using namespace std;
#define ll long long
int main(){
	ll N,D,A;
	cin >> N >> D >> A;
	ll i;
	queue< pair<ll,ll> > que;
	ll X,H;
	pair <ll , ll > enemy[N];
	ll tmp=0;
	ll ans=0;
	ll num;
	for(i=0;i<N;i++){
		cin >> X >>H;
		enemy[i]=make_pair(X,H);
	}
	sort(enemy,enemy+N);
	for(i=0;i<N;i++){
		while(!que.empty() && que.front().first < enemy[i].first){
			tmp-=que.front().second;
			que.pop();
		}
		if(enemy[i].second-tmp>0)num=(enemy[i].second-tmp-1)/A+1;else num=0;
		que.push(make_pair(enemy[i].first+2*D,num*A));
		tmp+=num*A;
		ans+=num;
	}
	cout << ans;
    return 0;
}

### Prompt

Generate 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<string>
#include<vector>
#include<iomanip>
#include<algorithm>
#include<queue>
#include<stack>
#include<map>
using namespace std;
#define ll long long
int main(){
	ll N,D,A;
	cin >> N >> D >> A;
	ll i;
	queue< pair<ll,ll> > que;
	ll X,H;
	pair <ll , ll > enemy[N];
	ll tmp=0;
	ll ans=0;
	ll num;
	for(i=0;i<N;i++){
		cin >> X >>H;
		enemy[i]=make_pair(X,H);
	}
	sort(enemy,enemy+N);
	for(i=0;i<N;i++){
		while(!que.empty() && que.front().first < enemy[i].first){
			tmp-=que.front().second;
			que.pop();
		}
		if(enemy[i].second-tmp>0)num=(enemy[i].second-tmp-1)/A+1;else num=0;
		que.push(make_pair(enemy[i].first+2*D,num*A));
		tmp+=num*A;
		ans+=num;
	}
	cout << ans;
    return 0;
}

```