#include<bits/stdc++.h>
#define ll long long
#define P pair<ll,ll>
using namespace std;

#define N 200000

int main(){
	int n;
	ll d;
	ll a;
	P monsters[N];
	scanf("%d%lld%lld",&n,&d,&a);
	d *= 2;
	for(int i=0;i<n;++i){
		scanf("%lld%lld",&monsters[i].first,&monsters[i].second);
	}
	sort(monsters,monsters+n);

	ll ans = 0;
	ll sum_damage = 0;
	int rp = 0;
	for(int i=0;i<n;++i){
		while(rp<n&&monsters[rp].first<=monsters[i].first+d){
			monsters[rp].second += sum_damage;
			++rp;
		}
		ll hp = monsters[i].second - sum_damage;
		if(hp>0){
			ll time = (hp + a - 1) / a;
			ans += time;
			sum_damage += time * a;
		}

	}

	printf("%lld\n",ans);

}

### 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>
#define ll long long
#define P pair<ll,ll>
using namespace std;

#define N 200000

int main(){
	int n;
	ll d;
	ll a;
	P monsters[N];
	scanf("%d%lld%lld",&n,&d,&a);
	d *= 2;
	for(int i=0;i<n;++i){
		scanf("%lld%lld",&monsters[i].first,&monsters[i].second);
	}
	sort(monsters,monsters+n);

	ll ans = 0;
	ll sum_damage = 0;
	int rp = 0;
	for(int i=0;i<n;++i){
		while(rp<n&&monsters[rp].first<=monsters[i].first+d){
			monsters[rp].second += sum_damage;
			++rp;
		}
		ll hp = monsters[i].second - sum_damage;
		if(hp>0){
			ll time = (hp + a - 1) / a;
			ans += time;
			sum_damage += time * a;
		}

	}

	printf("%lld\n",ans);

}

```