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

int main(){
  int N,D,A;
  cin>>N>>D>>A;
  vector<pair<int,int> > xh(N);
  for(int i=0;i<N;i++){
    int x,h;
    cin>>x>>h;
    xh.at(i)=make_pair(x,(h-1)/A+1);
  }

  sort(xh.begin(),xh.end());

  long long ans=0;
  vector<int> sum(N+1,0);
  for(int i=0;i<N;i++){
    if(i!=0){
      sum.at(i)+=sum.at(i-1);
    }
    if(sum.at(i)<xh.at(i).second){
      int tmp=xh.at(i).second-sum.at(i);
      ans+=tmp;
      int k=(int)(upper_bound(xh.begin(),xh.end(),make_pair(xh.at(i).first+2*D,1000000001))-xh.begin());
      sum.at(i)+=tmp;
      sum.at(k)-=tmp;
    }
  }

  cout<<ans<<endl;
}

### Prompt

Generate 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 N,D,A;
  cin>>N>>D>>A;
  vector<pair<int,int> > xh(N);
  for(int i=0;i<N;i++){
    int x,h;
    cin>>x>>h;
    xh.at(i)=make_pair(x,(h-1)/A+1);
  }

  sort(xh.begin(),xh.end());

  long long ans=0;
  vector<int> sum(N+1,0);
  for(int i=0;i<N;i++){
    if(i!=0){
      sum.at(i)+=sum.at(i-1);
    }
    if(sum.at(i)<xh.at(i).second){
      int tmp=xh.at(i).second-sum.at(i);
      ans+=tmp;
      int k=(int)(upper_bound(xh.begin(),xh.end(),make_pair(xh.at(i).first+2*D,1000000001))-xh.begin());
      sum.at(i)+=tmp;
      sum.at(k)-=tmp;
    }
  }

  cout<<ans<<endl;
}

```