#include<bits/stdc++.h>

using namespace std;

#define fi first
#define se second
#define ll long long


int main(){
	ll n,d,a;
	scanf("%lld%lld%lld",&n,&d,&a);
	vector<pair<ll,ll> > arr(n);
	for(int i=0;i<n;i++){
		scanf("%lld%lld",&arr[i].fi,&arr[i].se);
	}
	sort(arr.begin(),arr.end());
	vector<ll> cnt(n+1);
	ll i=0,j=0,res=0;
	while(i<n){
		if(arr[i].se>cnt[i]){
			ll v=arr[i].se-cnt[i];
			ll bomb=v/a;
			if(v%a!=0)
				bomb++;
			res+=bomb;
			ll dis=arr[i].fi+2*d;
			while(j<n&&arr[j].fi<=dis){
				j++;
			}
			cnt[i]+=(bomb*a);
			cnt[j]-=(bomb*a);
		}
		cnt[i+1]+=cnt[i];
		i++;
	}
	printf("%lld\n",res);
}

### 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;

#define fi first
#define se second
#define ll long long


int main(){
	ll n,d,a;
	scanf("%lld%lld%lld",&n,&d,&a);
	vector<pair<ll,ll> > arr(n);
	for(int i=0;i<n;i++){
		scanf("%lld%lld",&arr[i].fi,&arr[i].se);
	}
	sort(arr.begin(),arr.end());
	vector<ll> cnt(n+1);
	ll i=0,j=0,res=0;
	while(i<n){
		if(arr[i].se>cnt[i]){
			ll v=arr[i].se-cnt[i];
			ll bomb=v/a;
			if(v%a!=0)
				bomb++;
			res+=bomb;
			ll dis=arr[i].fi+2*d;
			while(j<n&&arr[j].fi<=dis){
				j++;
			}
			cnt[i]+=(bomb*a);
			cnt[j]-=(bomb*a);
		}
		cnt[i+1]+=cnt[i];
		i++;
	}
	printf("%lld\n",res);
}

```