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

#define i64 long long int
#define ran 202202

int n, D, A;
pair<int, int> a[ran];
i64 d[ran];

int main() {
	scanf("%d%d%d", &n, &D, &A);
	D*=2;
	for(int i=0;i<n;i++) {
		int x, h;
		scanf("%d%d", &x, &h);
		a[i] = make_pair(x, (h-1)/A+1);
	}
	sort(a, a+n);
	
	i64 done = 0LL, ans = 0LL;
	int it = 0;
	for(int i=0;i<n;i++){
		while(a[i].first - a[it].first > D) {
			done -= d[it];
			it ++;
		}
		
		i64 det = max(0LL, 1LL*a[i].second-done);
		done += det;
		d[i] = det;
		ans += det;
	}
	cout<<ans<<endl;
	
	return 0;
}
### Prompt

Develop a solution in cpp to the problem described below:

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 i64 long long int
#define ran 202202

int n, D, A;
pair<int, int> a[ran];
i64 d[ran];

int main() {
	scanf("%d%d%d", &n, &D, &A);
	D*=2;
	for(int i=0;i<n;i++) {
		int x, h;
		scanf("%d%d", &x, &h);
		a[i] = make_pair(x, (h-1)/A+1);
	}
	sort(a, a+n);
	
	i64 done = 0LL, ans = 0LL;
	int it = 0;
	for(int i=0;i<n;i++){
		while(a[i].first - a[it].first > D) {
			done -= d[it];
			it ++;
		}
		
		i64 det = max(0LL, 1LL*a[i].second-done);
		done += det;
		d[i] = det;
		ans += det;
	}
	cout<<ans<<endl;
	
	return 0;
}
```