#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N=4e5+10;
struct Node
{
	ll x,h;
}p[N];
ll n,d,a,sum,tmp[N];
bool Cmp(const Node& a,const Node& b)
{
	return a.x<b.x;
}
int main()
{
	scanf("%lld%lld%lld",&n,&d,&a);
	d=d*2+1;
	for (int i=1;i<=n;i++) scanf("%lld%lld",&p[i].x,&p[i].h);
	sort(p+1,p+1+n,Cmp);
	for (int i=1,j=1,now=0;i<=n;i++)
	{
		while (p[j+1].x<p[i].x+d&&j<n) j++;
		p[i].h-=now;
		if (p[i].h>=0) tmp[j]+=(p[i].h/a+(p[i].h%a!=0))*a, now+=(p[i].h/a+(p[i].h%a!=0))*a, sum+=p[i].h/a+(p[i].h%a!=0);
		now-=tmp[i];
	}
	printf("%lld\n",sum);
}
### 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<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N=4e5+10;
struct Node
{
	ll x,h;
}p[N];
ll n,d,a,sum,tmp[N];
bool Cmp(const Node& a,const Node& b)
{
	return a.x<b.x;
}
int main()
{
	scanf("%lld%lld%lld",&n,&d,&a);
	d=d*2+1;
	for (int i=1;i<=n;i++) scanf("%lld%lld",&p[i].x,&p[i].h);
	sort(p+1,p+1+n,Cmp);
	for (int i=1,j=1,now=0;i<=n;i++)
	{
		while (p[j+1].x<p[i].x+d&&j<n) j++;
		p[i].h-=now;
		if (p[i].h>=0) tmp[j]+=(p[i].h/a+(p[i].h%a!=0))*a, now+=(p[i].h/a+(p[i].h%a!=0))*a, sum+=p[i].h/a+(p[i].h%a!=0);
		now-=tmp[i];
	}
	printf("%lld\n",sum);
}
```