#include<cstdio>
#include<algorithm>
typedef long long ll;
using namespace std;
struct Z{
	ll x,h;
	int operator<(const Z&a)const{
		return x<a.x;
	}
}A[200001];
int i,j,n,d,a;
ll r,t,f[200001];
void add(int i,ll x){for(;i;i-=i&-i)f[i]+=x;}
ll val(int i){ll r=0;for(;i<=n;i+=i&-i)r+=f[i];return r;}
int main()
{
	scanf("%d%d%d",&n,&d,&a);
	for(;++i<=n;A[i].h=A[i].h/a+(A[i].h%a>0))scanf("%lld%lld",&A[i].x,&A[i].h);
	sort(A+1,A+n+1);
	for(i=1;i<=n;++i)if((t=A[i].h-val(i))>0)
	{
		j=upper_bound(A+1,A+n+1,(Z){A[i].x+d+d,0})-A-1;
		r+=t,add(j,t);
	}
	printf("%lld",r);
}
### 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<cstdio>
#include<algorithm>
typedef long long ll;
using namespace std;
struct Z{
	ll x,h;
	int operator<(const Z&a)const{
		return x<a.x;
	}
}A[200001];
int i,j,n,d,a;
ll r,t,f[200001];
void add(int i,ll x){for(;i;i-=i&-i)f[i]+=x;}
ll val(int i){ll r=0;for(;i<=n;i+=i&-i)r+=f[i];return r;}
int main()
{
	scanf("%d%d%d",&n,&d,&a);
	for(;++i<=n;A[i].h=A[i].h/a+(A[i].h%a>0))scanf("%lld%lld",&A[i].x,&A[i].h);
	sort(A+1,A+n+1);
	for(i=1;i<=n;++i)if((t=A[i].h-val(i))>0)
	{
		j=upper_bound(A+1,A+n+1,(Z){A[i].x+d+d,0})-A-1;
		r+=t,add(j,t);
	}
	printf("%lld",r);
}
```