#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 5;
int n, d, a;
pair<int, int> p[maxn];
int main() {
	scanf("%d%d%d", &n, &d, &a);
	for(int i = 1; i <= n; ++i) {
		scanf("%d%d", &p[i].first, &p[i].second);
		p[i].second = (p[i].second + a - 1) / a;
	}
	sort(p + 1, p + n + 1);
	int cnt = 0;
	long long ans = 0;
	for(int i = 1, j = 1; i <= n; ++i) {
		while(p[i].first - p[j].first > 2 * d) {
			cnt -= p[j].second;
			++j;
		}
		ans += max(0, p[i].second - cnt);
		p[i].second = max(0, p[i].second - cnt);
		cnt += p[i].second;
	}
	printf("%lld\n", ans);
	return 0;
}
### Prompt

Please create a solution in Cpp 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;
const int maxn = 2e5 + 5;
int n, d, a;
pair<int, int> p[maxn];
int main() {
	scanf("%d%d%d", &n, &d, &a);
	for(int i = 1; i <= n; ++i) {
		scanf("%d%d", &p[i].first, &p[i].second);
		p[i].second = (p[i].second + a - 1) / a;
	}
	sort(p + 1, p + n + 1);
	int cnt = 0;
	long long ans = 0;
	for(int i = 1, j = 1; i <= n; ++i) {
		while(p[i].first - p[j].first > 2 * d) {
			cnt -= p[j].second;
			++j;
		}
		ans += max(0, p[i].second - cnt);
		p[i].second = max(0, p[i].second - cnt);
		cnt += p[i].second;
	}
	printf("%lld\n", ans);
	return 0;
}
```