#include <bits/stdc++.h>
#define MOD 1000000007LL
using namespace std;
typedef long long ll;
typedef pair<int,int> P;

int n;
int d,a;
P p[200005];
ll sum[200005];

int main(void){
	scanf("%d%d%d",&n,&d,&a);
	for(int i=0;i<n;i++){
		scanf("%d%d",&p[i].first,&p[i].second);
	}
	sort(p,p+n);
	ll ans=0;
	for(int i=0;i<n;i++){
		ll rest=p[i].second-sum[i];
		if(rest<=0){
			sum[i+1]+=sum[i];
			continue;
		}
		ll cnt=(rest+a-1LL)/a;
		ans+=cnt;
		sum[i+1]+=sum[i]+cnt*a;
		int pos=lower_bound(p,p+n,P(p[i].first+2LL*d+1LL,0))-p;
		sum[pos]-=cnt*a;
	}
	printf("%lld\n",ans);
	return 0;
}

### Prompt

Create a solution in CPP for 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>
#define MOD 1000000007LL
using namespace std;
typedef long long ll;
typedef pair<int,int> P;

int n;
int d,a;
P p[200005];
ll sum[200005];

int main(void){
	scanf("%d%d%d",&n,&d,&a);
	for(int i=0;i<n;i++){
		scanf("%d%d",&p[i].first,&p[i].second);
	}
	sort(p,p+n);
	ll ans=0;
	for(int i=0;i<n;i++){
		ll rest=p[i].second-sum[i];
		if(rest<=0){
			sum[i+1]+=sum[i];
			continue;
		}
		ll cnt=(rest+a-1LL)/a;
		ans+=cnt;
		sum[i+1]+=sum[i]+cnt*a;
		int pos=lower_bound(p,p+n,P(p[i].first+2LL*d+1LL,0))-p;
		sum[pos]-=cnt*a;
	}
	printf("%lld\n",ans);
	return 0;
}

```