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

using ll = long long;
using  P = pair<int, int>;

int main() {
	int n; ll d, a;
	cin >> n >> d >> a;
	vector<P>p(n);
	for (int i = 0; i < n; i++) {
		cin >> p[i].first >> p[i].second;
	}
	sort(p.begin(), p.end());
	d = d * 2;
	ll tot = 0;
	queue<pair<ll, ll>>q;
	ll ans = 0;
	for (int i = 0; i < n; i++) {
		ll x = p[i].first;
		ll h = p[i].second;
		while (q.size() && q.front().first < x) {
			tot -= q.front().second;
			q.pop();
		}
		h -= tot;
		if (h > 0) {
			ll num = (h + a - 1) / a;
			ans += num;
			ll damage = num * a;
			tot += damage;
			q.emplace(x + d, damage);
		}
	}
	cout << ans << endl;
	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<iostream>
#include<vector>
#include<algorithm>
#include<queue>
using namespace std;

using ll = long long;
using  P = pair<int, int>;

int main() {
	int n; ll d, a;
	cin >> n >> d >> a;
	vector<P>p(n);
	for (int i = 0; i < n; i++) {
		cin >> p[i].first >> p[i].second;
	}
	sort(p.begin(), p.end());
	d = d * 2;
	ll tot = 0;
	queue<pair<ll, ll>>q;
	ll ans = 0;
	for (int i = 0; i < n; i++) {
		ll x = p[i].first;
		ll h = p[i].second;
		while (q.size() && q.front().first < x) {
			tot -= q.front().second;
			q.pop();
		}
		h -= tot;
		if (h > 0) {
			ll num = (h + a - 1) / a;
			ans += num;
			ll damage = num * a;
			tot += damage;
			q.emplace(x + d, damage);
		}
	}
	cout << ans << endl;
	return 0;
}
```