#include <bits/stdc++.h>
#define rep(i,n) for(int i=0; i<(int)(n); i++)

using namespace std;
typedef long long LL;
typedef pair<LL,LL> P;
const LL INF=1e12;

int main(){
	int N, D, A;
	cin >> N >> D >> A;
	vector<P> xh(N);
	rep(i,N){
		LL x, h;
		cin >> x >> h;
		xh[i]=P(x,(h+A-1)/A);
	}
	sort(xh.begin(),xh.end());
	vector<int> reach(N+1), damage(N+1);
	rep(i,N){
		P p=make_pair(xh[i].first+2*D,INF);
		int r=upper_bound(xh.begin(),xh.end(),p)-xh.begin();
		reach[i]=r;
	}
	LL ans=0;
	rep(i,N){
		if(i!=0) damage[i]+=damage[i-1];
		if(xh[i].second<=damage[i]) continue;
		LL num=xh[i].second-damage[i];
		ans+=num;
		damage[i]+=num, damage[reach[i]]-=num;
	}
	cout << ans << endl;

	return 0;
}
### Prompt

Your challenge is to write 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>
#define rep(i,n) for(int i=0; i<(int)(n); i++)

using namespace std;
typedef long long LL;
typedef pair<LL,LL> P;
const LL INF=1e12;

int main(){
	int N, D, A;
	cin >> N >> D >> A;
	vector<P> xh(N);
	rep(i,N){
		LL x, h;
		cin >> x >> h;
		xh[i]=P(x,(h+A-1)/A);
	}
	sort(xh.begin(),xh.end());
	vector<int> reach(N+1), damage(N+1);
	rep(i,N){
		P p=make_pair(xh[i].first+2*D,INF);
		int r=upper_bound(xh.begin(),xh.end(),p)-xh.begin();
		reach[i]=r;
	}
	LL ans=0;
	rep(i,N){
		if(i!=0) damage[i]+=damage[i-1];
		if(xh[i].second<=damage[i]) continue;
		LL num=xh[i].second-damage[i];
		ans+=num;
		damage[i]+=num, damage[reach[i]]-=num;
	}
	cout << ans << endl;

	return 0;
}
```