#include <cstdio>
#include <queue>
#include <algorithm>

using namespace std;

struct node {
	long long X, H;
	bool operator < (const node& x) const {
		return X < x.X;
	}
}a[200000];
queue<node> h;
long long N, D, A, s, ans;

int main() {
	scanf("%lld%lld%lld", &N, &D, &A);
	D = 2 * D + 1;
	for (int i = 0; i < N; ++i) scanf("%lld%lld", &a[i].X, &a[i].H);
	sort(a, a + N);
	for (int i = 0; i < N; ++i) {
		while (!h.empty() && h.front().X <= a[i].X) {
			s -= h.front().H;
			h.pop();
		}
		if (s < a[i].H) {
			long long d = (a[i].H - s - 1) / A + 1;
			h.push({a[i].X + D, A * d});
			ans += d;
			s += A * d;
		}
	}
	printf("%lld\n", ans);
	return 0;
}
### Prompt

Please provide a CPP coded solution 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 <cstdio>
#include <queue>
#include <algorithm>

using namespace std;

struct node {
	long long X, H;
	bool operator < (const node& x) const {
		return X < x.X;
	}
}a[200000];
queue<node> h;
long long N, D, A, s, ans;

int main() {
	scanf("%lld%lld%lld", &N, &D, &A);
	D = 2 * D + 1;
	for (int i = 0; i < N; ++i) scanf("%lld%lld", &a[i].X, &a[i].H);
	sort(a, a + N);
	for (int i = 0; i < N; ++i) {
		while (!h.empty() && h.front().X <= a[i].X) {
			s -= h.front().H;
			h.pop();
		}
		if (s < a[i].H) {
			long long d = (a[i].H - s - 1) / A + 1;
			h.push({a[i].X + D, A * d});
			ans += d;
			s += A * d;
		}
	}
	printf("%lld\n", ans);
	return 0;
}
```