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

signed main(){
	int N, D, A, X, H;
	scanf("%lld%lld%lld", &N, &D, &A);
	priority_queue<pair<int, int>, vector<pair<int, int>>, greater<pair<int, int>>> pq;
	while(N--){
		scanf("%lld%lld", &X, &H);
		pq.push(make_pair(X, H));
	}
	int ans = 0, now = 0;
	while(pq.size() > 0){
		X = pq.top().first;
		H = pq.top().second;
		pq.pop();
//		printf("(X, H) = (%lld, %lld)\n", X, H);
		if(H < 0){
			now += H;
		}
		else{
			if(H > now){
				ans += (H - now + A - 1) / A;
				pq.push(make_pair(X + 2 * D + 1, -((H - now + A - 1) / A) * A));
				now += (H - now + A - 1) / A * A;
			}
		}
	}
	printf("%lld\n", ans);
	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>
using namespace std;
#define int long long

signed main(){
	int N, D, A, X, H;
	scanf("%lld%lld%lld", &N, &D, &A);
	priority_queue<pair<int, int>, vector<pair<int, int>>, greater<pair<int, int>>> pq;
	while(N--){
		scanf("%lld%lld", &X, &H);
		pq.push(make_pair(X, H));
	}
	int ans = 0, now = 0;
	while(pq.size() > 0){
		X = pq.top().first;
		H = pq.top().second;
		pq.pop();
//		printf("(X, H) = (%lld, %lld)\n", X, H);
		if(H < 0){
			now += H;
		}
		else{
			if(H > now){
				ans += (H - now + A - 1) / A;
				pq.push(make_pair(X + 2 * D + 1, -((H - now + A - 1) / A) * A));
				now += (H - now + A - 1) / A * A;
			}
		}
	}
	printf("%lld\n", ans);
	return 0;
}
```