#include <bits/stdc++.h> 
#define ll long long
using namespace std;

ll n,d,a,j=1ll,sumv=0ll,total=0ll;
ll bomb[200005];

pair<ll,ll>m[200005];

int main()
{
	cin>>n>>d>>a;
	for (int i=1;i<=n;i++)
	{
		cin>>m[i].first>>m[i].second;
		m[i].second=(m[i].second+a-1)/a;
	}
	sort(m+1,m+n+1);
	for (int i=1;i<=n;i++)
	{
		while ((j<i)&&(m[i].first-m[j].first>d*2))
		{
			sumv-=bomb[j];
			j+=1ll;
		}
		bomb[i]=max(m[i].second-sumv,0ll);
		sumv+=bomb[i];
		total+=bomb[i];
	}
	cout<<total<<endl;
	return 0;
}
### Prompt

Please create a solution in cpp 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> 
#define ll long long
using namespace std;

ll n,d,a,j=1ll,sumv=0ll,total=0ll;
ll bomb[200005];

pair<ll,ll>m[200005];

int main()
{
	cin>>n>>d>>a;
	for (int i=1;i<=n;i++)
	{
		cin>>m[i].first>>m[i].second;
		m[i].second=(m[i].second+a-1)/a;
	}
	sort(m+1,m+n+1);
	for (int i=1;i<=n;i++)
	{
		while ((j<i)&&(m[i].first-m[j].first>d*2))
		{
			sumv-=bomb[j];
			j+=1ll;
		}
		bomb[i]=max(m[i].second-sumv,0ll);
		sumv+=bomb[i];
		total+=bomb[i];
	}
	cout<<total<<endl;
	return 0;
}
```