#include<bits/stdc++.h>
using namespace std;
#define int long long
struct node
{
	int x,h;
}t[1000001];
bool cmp(node a,node b){return a.x<b.x;}
signed main()
{
	int n,d,a;
	scanf("%lld%lld%lld",&n,&d,&a);
	for(int i=1;i<=n;i++)
		scanf("%lld%lld",&t[i].x,&t[i].h);
	sort(t+1,t+n+1,cmp);
	long long add=0,ans=0;
	priority_queue<pair<int,int> ,vector<pair<int,int> >,greater<pair<int,int> > >q;
	for(int i=1;i<=n;i++)
	{
		while(!q.empty()&&t[i].x-q.top().first>2*d)
		{
			add-=1ll*q.top().second*a;
			q.pop();
		}
		int delta=t[i].h-add;
		if(delta>0)
		{
			int tmp=(delta+a-1)/a;
			ans+=tmp;
			add+=tmp*a;
			q.push(make_pair(t[i].x,tmp));
		}
	}
	printf("%lld",ans);
	return 0;
}

### Prompt

Create a solution in Cpp for 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;
#define int long long
struct node
{
	int x,h;
}t[1000001];
bool cmp(node a,node b){return a.x<b.x;}
signed main()
{
	int n,d,a;
	scanf("%lld%lld%lld",&n,&d,&a);
	for(int i=1;i<=n;i++)
		scanf("%lld%lld",&t[i].x,&t[i].h);
	sort(t+1,t+n+1,cmp);
	long long add=0,ans=0;
	priority_queue<pair<int,int> ,vector<pair<int,int> >,greater<pair<int,int> > >q;
	for(int i=1;i<=n;i++)
	{
		while(!q.empty()&&t[i].x-q.top().first>2*d)
		{
			add-=1ll*q.top().second*a;
			q.pop();
		}
		int delta=t[i].h-add;
		if(delta>0)
		{
			int tmp=(delta+a-1)/a;
			ans+=tmp;
			add+=tmp*a;
			q.push(make_pair(t[i].x,tmp));
		}
	}
	printf("%lld",ans);
	return 0;
}

```