#include<bits/stdc++.h>
using namespace std;
#define LL long long
int n,a;
LL d;
struct data{
	LL h,x;
}w[200005];
LL ans;
queue<data>Q;
bool cmp(data a,data b){
	return a.x<b.x;
}
int main(){
	scanf("%d%lld%d",&n,&d,&a);
	for(int i=1;i<=n;++i){
		scanf("%lld%lld",&w[i].x,&w[i].h);
	}
	sort(w+1,w+1+n,cmp);
	LL now=0;
	for(int i=1;i<=n;++i){
		while(!Q.empty()&&Q.front().x<w[i].x){now-=Q.front().h;Q.pop();}
		data e;
		e.x=w[i].x+d*2ll;
		w[i].h-=now;
		if(w[i].h<=0)continue;
		e.h=(w[i].h/a+(w[i].h%a!=0))*a;
		now+=e.h;
		ans+=e.h/a;
		Q.push(e);
	}
	printf("%lld\n",ans);
	return 0;
}
### 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>
using namespace std;
#define LL long long
int n,a;
LL d;
struct data{
	LL h,x;
}w[200005];
LL ans;
queue<data>Q;
bool cmp(data a,data b){
	return a.x<b.x;
}
int main(){
	scanf("%d%lld%d",&n,&d,&a);
	for(int i=1;i<=n;++i){
		scanf("%lld%lld",&w[i].x,&w[i].h);
	}
	sort(w+1,w+1+n,cmp);
	LL now=0;
	for(int i=1;i<=n;++i){
		while(!Q.empty()&&Q.front().x<w[i].x){now-=Q.front().h;Q.pop();}
		data e;
		e.x=w[i].x+d*2ll;
		w[i].h-=now;
		if(w[i].h<=0)continue;
		e.h=(w[i].h/a+(w[i].h%a!=0))*a;
		now+=e.h;
		ans+=e.h/a;
		Q.push(e);
	}
	printf("%lld\n",ans);
	return 0;
}
```