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


pair <int,int> mon[200005];


int main(){
	ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
	int n, d, a;
	cin >> n >> d >> a;
	for (int i = 1; i <= n; ++i){
		int x, h;
		cin >> x >> h;
		mon[i] = {x,h};
	}
	sort(mon+1,mon+n+1);
	queue <pair<int,int>> B;
	long long akt_dmg = 0;
	long long wynik = 0;

	for (int i = 1; i <= n; ++i){
		int x = mon[i].first, h = mon[i].second;
		while (!B.empty() && B.front().first + d < x){
			akt_dmg -= B.front().second;
			B.pop();
		}
		h -= akt_dmg;
		if (h > 0){
			int ile = h / a + (h % a != 0);
			B.push({x+d,ile*a});
			akt_dmg += ile*a;
			wynik += ile;
		}
	}
	cout << wynik << "\n";
}
### Prompt

Construct a CPP code solution to the problem outlined:

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;


pair <int,int> mon[200005];


int main(){
	ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
	int n, d, a;
	cin >> n >> d >> a;
	for (int i = 1; i <= n; ++i){
		int x, h;
		cin >> x >> h;
		mon[i] = {x,h};
	}
	sort(mon+1,mon+n+1);
	queue <pair<int,int>> B;
	long long akt_dmg = 0;
	long long wynik = 0;

	for (int i = 1; i <= n; ++i){
		int x = mon[i].first, h = mon[i].second;
		while (!B.empty() && B.front().first + d < x){
			akt_dmg -= B.front().second;
			B.pop();
		}
		h -= akt_dmg;
		if (h > 0){
			int ile = h / a + (h % a != 0);
			B.push({x+d,ile*a});
			akt_dmg += ile*a;
			wynik += ile;
		}
	}
	cout << wynik << "\n";
}
```