#include<bits/stdc++.h>
#define long long long
using namespace std;
const long N=1e6,inf=1e9;
pair<long,long>pp[N+2];
int main()
{
  ios_base::sync_with_stdio(0);cin.tie(0);
  long n,d,a;cin>>n>>d>>a;

  for(long i=1;i<=n;i++)
    cin>>pp[i].first>>pp[i].second;
  sort(pp+1,pp+n+1);

  priority_queue<pair<long,long> >pq;long cnt=0;
  long ans=0;
  for(long i=1;i<=n;i++)
  {
    while(cnt)
    {
      if((-pq.top().first)<pp[i].first)
      {
        cnt-=pq.top().second;pq.pop();
      }
      else break;
    }
    long h=pp[i].second-cnt*a;
    if(h<=0)continue;
    long ad=(h+a-1)/a;pq.push({-(pp[i].first+2*d),ad});
    cnt+=ad;ans+=ad;
  }
  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>
#define long long long
using namespace std;
const long N=1e6,inf=1e9;
pair<long,long>pp[N+2];
int main()
{
  ios_base::sync_with_stdio(0);cin.tie(0);
  long n,d,a;cin>>n>>d>>a;

  for(long i=1;i<=n;i++)
    cin>>pp[i].first>>pp[i].second;
  sort(pp+1,pp+n+1);

  priority_queue<pair<long,long> >pq;long cnt=0;
  long ans=0;
  for(long i=1;i<=n;i++)
  {
    while(cnt)
    {
      if((-pq.top().first)<pp[i].first)
      {
        cnt-=pq.top().second;pq.pop();
      }
      else break;
    }
    long h=pp[i].second-cnt*a;
    if(h<=0)continue;
    long ad=(h+a-1)/a;pq.push({-(pp[i].first+2*d),ad});
    cnt+=ad;ans+=ad;
  }
  cout<<ans<<endl;
  return 0;
}
```