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

int main(int argc, char const *argv[])
{
	int n;
	long long d,a;
	cin>>n>>d>>a;
	pair<long long,long long> xh[n];
	for (int i = 0; i < n; ++i)
	{
		cin>>xh[i].first>>xh[i].second;
	}
	sort(xh,xh+n);
	long long ans=0LL,now=0LL;
	pair<long long,long long> bomb[n];
	int r=0,w=0;
	for (int i = 0; i < n; ++i)
	{
		while (i>0&&xh[i].first>bomb[r].first)
		{
			now-=bomb[r].second;
			++r;
			if (r>=w)
			{
				break;
			}
		}
		if (xh[i].second-now>0LL)
		{
			long long b=(xh[i].second-now+a-1LL)/a;
			ans+=b;
			now+=a*b;
			bomb[w]=make_pair(xh[i].first+2LL*d,a*b);
			++w;
			int j=i;
		}
	}
	cout<<ans<<endl;
	return 0;
}

### Prompt

Please formulate 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;

int main(int argc, char const *argv[])
{
	int n;
	long long d,a;
	cin>>n>>d>>a;
	pair<long long,long long> xh[n];
	for (int i = 0; i < n; ++i)
	{
		cin>>xh[i].first>>xh[i].second;
	}
	sort(xh,xh+n);
	long long ans=0LL,now=0LL;
	pair<long long,long long> bomb[n];
	int r=0,w=0;
	for (int i = 0; i < n; ++i)
	{
		while (i>0&&xh[i].first>bomb[r].first)
		{
			now-=bomb[r].second;
			++r;
			if (r>=w)
			{
				break;
			}
		}
		if (xh[i].second-now>0LL)
		{
			long long b=(xh[i].second-now+a-1LL)/a;
			ans+=b;
			now+=a*b;
			bomb[w]=make_pair(xh[i].first+2LL*d,a*b);
			++w;
			int j=i;
		}
	}
	cout<<ans<<endl;
	return 0;
}

```