#include <bits/stdc++.h>

using namespace std;

bool cmp(vector<int>& x, vector<int>& y) {
	return x[0] < y[0];
}

int main() {
	int n, d;
	long long a;
	cin >> n >> d >> a;
	vector<vector<int>> e(n, vector<int>(2));
	for (int i = 0; i < n; i++) {
		cin >> e[i][0] >> e[i][1];
	}
	sort(e.begin(), e.end(), cmp);
	queue<pair<int, int>> b;
	long long c = 0, ans = 0;
	for (int i = 0; i < n; i++) {
		while (b.size() > 0 && b.front().first < e[i][0] - d) {
			c -= b.front().second;
			b.pop();
		}
		int t = (max(1LL * e[i][1] - a * c, 0LL) + a - 1) / a;
		b.push(make_pair(e[i][0] + d, t));
		c += t;
		ans += t;
	}
	cout << ans << endl;
	return 0;
}
### Prompt

In CPP, your task is to solve 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;

bool cmp(vector<int>& x, vector<int>& y) {
	return x[0] < y[0];
}

int main() {
	int n, d;
	long long a;
	cin >> n >> d >> a;
	vector<vector<int>> e(n, vector<int>(2));
	for (int i = 0; i < n; i++) {
		cin >> e[i][0] >> e[i][1];
	}
	sort(e.begin(), e.end(), cmp);
	queue<pair<int, int>> b;
	long long c = 0, ans = 0;
	for (int i = 0; i < n; i++) {
		while (b.size() > 0 && b.front().first < e[i][0] - d) {
			c -= b.front().second;
			b.pop();
		}
		int t = (max(1LL * e[i][1] - a * c, 0LL) + a - 1) / a;
		b.push(make_pair(e[i][0] + d, t));
		c += t;
		ans += t;
	}
	cout << ans << endl;
	return 0;
}
```