#include <iostream>
#include <vector>
using namespace std;

int main() {
	long long n, d, a;
	cin >> n >> d >> a;
	vector<pair<long long, long long>> x(n);
	for (int i = 0; i < n; i++) {
		cin >> x[i].first >> x[i].second;
		x[i].second = (x[i].second + a - 1) / a;
	}
	sort(x.begin(), x.end());
	long long ans = 0;
	vector<long long> v(n+1, 0);
	for (int i = 0, j = 0; i < n; i++) {
		if (v[i] < x[i].second) {
			long long b = x[i].second - v[i];
			ans += b;
			v[i] += b;
			while (j < n && x[j].first <= x[i].first + 2*d) j++;
			v[j] -= b;
		}
		v[i+1] += v[i];
	}
	cout << ans << endl;
	return 0;
}
### Prompt

Generate 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 <vector>
using namespace std;

int main() {
	long long n, d, a;
	cin >> n >> d >> a;
	vector<pair<long long, long long>> x(n);
	for (int i = 0; i < n; i++) {
		cin >> x[i].first >> x[i].second;
		x[i].second = (x[i].second + a - 1) / a;
	}
	sort(x.begin(), x.end());
	long long ans = 0;
	vector<long long> v(n+1, 0);
	for (int i = 0, j = 0; i < n; i++) {
		if (v[i] < x[i].second) {
			long long b = x[i].second - v[i];
			ans += b;
			v[i] += b;
			while (j < n && x[j].first <= x[i].first + 2*d) j++;
			v[j] -= b;
		}
		v[i+1] += v[i];
	}
	cout << ans << endl;
	return 0;
}
```