#include <bits/stdc++.h>
using namespace std;

int main(){
	ios::sync_with_stdio(0);
	cout << setprecision(10) << fixed;
	long long n, d, a;
	cin >> n >> d >> a;
	vector<pair<long long, long long>> m(n);
	vector<long long> used(n + 1, 0);
	for (int i = 0; i < n; ++i){
		cin >> m[i].first >> m[i].second;
	}
	sort(m.begin(), m.end());
	long long ans = 0;
	for (int i = 0, j = 0; i < n; ++i){
		while (j < n && m[i].first + 2 * d >= m[j].first){
			j++;
		}
		int need = max((1ll * m[i].second - used[i] * a + a - 1) / a, 0ll);
		ans += need;
		used[i] += need;
		used[j] -= need;
		used[i + 1] += used[i];
	}
	cout << ans;
	return 0;
}


### Prompt

Develop a solution in Cpp 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 <bits/stdc++.h>
using namespace std;

int main(){
	ios::sync_with_stdio(0);
	cout << setprecision(10) << fixed;
	long long n, d, a;
	cin >> n >> d >> a;
	vector<pair<long long, long long>> m(n);
	vector<long long> used(n + 1, 0);
	for (int i = 0; i < n; ++i){
		cin >> m[i].first >> m[i].second;
	}
	sort(m.begin(), m.end());
	long long ans = 0;
	for (int i = 0, j = 0; i < n; ++i){
		while (j < n && m[i].first + 2 * d >= m[j].first){
			j++;
		}
		int need = max((1ll * m[i].second - used[i] * a + a - 1) / a, 0ll);
		ans += need;
		used[i] += need;
		used[j] -= need;
		used[i + 1] += used[i];
	}
	cout << ans;
	return 0;
}


```