#include<bits/stdc++.h>
using namespace std;
#define rep(i,n) for(int i=0; i<(n); i++)
using ll = long long;
using P = pair<ll, ll>;
constexpr int MX = 200005;
int n,d,a,x,h;
long long ans,ac;
queue<P> qu;
P p[MX];
int main(){
	scanf("%d%d%d", &n, &d, &a);
	rep(i,n){
		scanf("%d%d", &x, &h);
		p[i] = P{x,h};
	}
	sort(p,p+n);
	rep(i,n){
		while(!qu.empty() && qu.front().first < p[i].first){
			ac -= qu.front().second;
			qu.pop();
		}
		if(ac < p[i].second){
			ll t = (p[i].second - ac - 1) / a + 1;
			ans += t;
			ac += a * t;
			qu.push(P{p[i].first+d*2,a*t});
		}
	}
	printf("%lld\n", ans);
}
### Prompt

Your challenge is to write 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;
#define rep(i,n) for(int i=0; i<(n); i++)
using ll = long long;
using P = pair<ll, ll>;
constexpr int MX = 200005;
int n,d,a,x,h;
long long ans,ac;
queue<P> qu;
P p[MX];
int main(){
	scanf("%d%d%d", &n, &d, &a);
	rep(i,n){
		scanf("%d%d", &x, &h);
		p[i] = P{x,h};
	}
	sort(p,p+n);
	rep(i,n){
		while(!qu.empty() && qu.front().first < p[i].first){
			ac -= qu.front().second;
			qu.pop();
		}
		if(ac < p[i].second){
			ll t = (p[i].second - ac - 1) / a + 1;
			ans += t;
			ac += a * t;
			qu.push(P{p[i].first+d*2,a*t});
		}
	}
	printf("%lld\n", ans);
}
```