#include <iostream>
#include <algorithm>
#include <cstdlib>
#include <cstdio>
#include <vector>

#define LL long long

using namespace std;

const int N = 200010;
int n, D, A;

struct T {
	int x, h;
}a[N];

LL tot[N];

int main() {
	scanf("%d%d%d", &n, &D, &A);

	for (int i = 1; i <= n; i++)
		scanf("%d%d", &a[i].x, &a[i].h);

	sort(a + 1, a + 1 + n, [](const T &p, const T &q) {
		return p.x < q.x;
	});

	LL ans = 0, sum = 0;
	for (int i = 1, j = 1; i <= n; i++) {
		while (j <= n && a[j].x - a[i].x <= 2 * D)
			j++;

		sum += tot[i];
		LL t = a[i].h - sum * A;
		if (t > 0) {
			LL c = t / A + (bool)(t % A);
			ans += c;

			sum += c;
			tot[j] -= c;
		}
	}

	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 <iostream>
#include <algorithm>
#include <cstdlib>
#include <cstdio>
#include <vector>

#define LL long long

using namespace std;

const int N = 200010;
int n, D, A;

struct T {
	int x, h;
}a[N];

LL tot[N];

int main() {
	scanf("%d%d%d", &n, &D, &A);

	for (int i = 1; i <= n; i++)
		scanf("%d%d", &a[i].x, &a[i].h);

	sort(a + 1, a + 1 + n, [](const T &p, const T &q) {
		return p.x < q.x;
	});

	LL ans = 0, sum = 0;
	for (int i = 1, j = 1; i <= n; i++) {
		while (j <= n && a[j].x - a[i].x <= 2 * D)
			j++;

		sum += tot[i];
		LL t = a[i].h - sum * A;
		if (t > 0) {
			LL c = t / A + (bool)(t % A);
			ans += c;

			sum += c;
			tot[j] -= c;
		}
	}

	printf("%lld\n", ans);

	return 0;
}
```