#include"bits/stdc++.h"
using namespace std;
typedef long long ll;
#define int ll
const int inf=1e18;
int32_t main()
{
	#ifndef ONLINE_JUDGE
		freopen("in","r", stdin);
		freopen("out","w",stdout);
	#endif

	int n,d,a;
	cin>>n>>d>>a;
	vector<pair<int,int>> p(n);
	for(int i=0;i<n;i++)
	{
		cin>>p[i].first>>p[i].second;
	}
	sort(p.begin(),p.end());
	int sum=0,res=0;
	queue<pair<int,int>> q;
	for(int i=0;i<n;i++)
	{
		while(!q.empty() and q.front().first<p[i].first)
		{
			sum-=q.front().second;
			q.pop();
		}

		p[i].second-=a*sum;
		if(p[i].second<=0)continue;
		int tmp=(p[i].second+a-1)/a;
		res+=tmp;
		q.push({p[i].first+2*d,tmp});
		sum+=tmp;
	}
	cout<<res<<'\n';


}
### Prompt

Your task is to create 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;
#define int ll
const int inf=1e18;
int32_t main()
{
	#ifndef ONLINE_JUDGE
		freopen("in","r", stdin);
		freopen("out","w",stdout);
	#endif

	int n,d,a;
	cin>>n>>d>>a;
	vector<pair<int,int>> p(n);
	for(int i=0;i<n;i++)
	{
		cin>>p[i].first>>p[i].second;
	}
	sort(p.begin(),p.end());
	int sum=0,res=0;
	queue<pair<int,int>> q;
	for(int i=0;i<n;i++)
	{
		while(!q.empty() and q.front().first<p[i].first)
		{
			sum-=q.front().second;
			q.pop();
		}

		p[i].second-=a*sum;
		if(p[i].second<=0)continue;
		int tmp=(p[i].second+a-1)/a;
		res+=tmp;
		q.push({p[i].first+2*d,tmp});
		sum+=tmp;
	}
	cout<<res<<'\n';


}
```