#include <bits/stdc++.h>
 
using namespace std;
 
int main() {
	ios_base::sync_with_stdio(0);
	cin.tie(NULL);
 
	int n;
	long long d, a;
	cin >> n >> d >> a;
	vector<pair<long long, long long>> v(n);
	for(int i = 0; i < n; i++) {
		cin >> v[i].first >> v[i].second;
	}
	sort(v.begin(), v.end());
	queue<long long> q, p;
	long long cur = 0;
	long long ans = 0;
	for(auto i : v) {
		while(!q.empty() and q.front() < i.first - 2*d) {
			q.pop();
			cur -= p.front();
			p.pop();
		}
		if(i.second <= cur) continue;
		long long req = (i.second - cur)/a;
		if((i.second - cur) % a != 0) req++;
		cur += req * a;
		ans += req;
		q.push(i.first);
		p.push(req*a);
	}
	cout << ans << '\n';
 
	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 <bits/stdc++.h>
 
using namespace std;
 
int main() {
	ios_base::sync_with_stdio(0);
	cin.tie(NULL);
 
	int n;
	long long d, a;
	cin >> n >> d >> a;
	vector<pair<long long, long long>> v(n);
	for(int i = 0; i < n; i++) {
		cin >> v[i].first >> v[i].second;
	}
	sort(v.begin(), v.end());
	queue<long long> q, p;
	long long cur = 0;
	long long ans = 0;
	for(auto i : v) {
		while(!q.empty() and q.front() < i.first - 2*d) {
			q.pop();
			cur -= p.front();
			p.pop();
		}
		if(i.second <= cur) continue;
		long long req = (i.second - cur)/a;
		if((i.second - cur) % a != 0) req++;
		cur += req * a;
		ans += req;
		q.push(i.first);
		p.push(req*a);
	}
	cout << ans << '\n';
 
	return 0;
}
```