//差分
#include <bits/stdc++.h>
using namespace std;
const int M=2e5+10;
typedef long long ll;
struct node{
	ll x,h;
}a[M];
ll c[M];

bool cmp(node a,node b){
	return a.x<b.x;
}

int main(int argc, char const *argv[])
{
	int N,D,A;
	cin>>N>>D>>A;
	for(int i=0;i<N;i++)scanf("%lld%lld",&a[i].x,&a[i].h);
	sort(a,a+N,cmp);
	
	ll ans=0;
	for(int i=0,j=0;i<N;i++){
		while(j<N&&a[j].x<=a[i].x+2*D)j++;
		ll tmp=max((a[i].h-c[i]*A+A-1)/A,0ll);
		c[i]+=tmp;
		c[j]-=tmp;//保证攻击i点的时候，j点没有攻击
		ans+=tmp;
		c[i+1]+=c[i];
	}
	cout<<ans<<endl;
	return 0;
}


### 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
//差分
#include <bits/stdc++.h>
using namespace std;
const int M=2e5+10;
typedef long long ll;
struct node{
	ll x,h;
}a[M];
ll c[M];

bool cmp(node a,node b){
	return a.x<b.x;
}

int main(int argc, char const *argv[])
{
	int N,D,A;
	cin>>N>>D>>A;
	for(int i=0;i<N;i++)scanf("%lld%lld",&a[i].x,&a[i].h);
	sort(a,a+N,cmp);
	
	ll ans=0;
	for(int i=0,j=0;i<N;i++){
		while(j<N&&a[j].x<=a[i].x+2*D)j++;
		ll tmp=max((a[i].h-c[i]*A+A-1)/A,0ll);
		c[i]+=tmp;
		c[j]-=tmp;//保证攻击i点的时候，j点没有攻击
		ans+=tmp;
		c[i+1]+=c[i];
	}
	cout<<ans<<endl;
	return 0;
}


```