#include<bits/stdc++.h>
using namespace std;
int main(){
  int n,d,a;
  cin>>n>>d>>a;
  vector<vector<int>> xh(n,vector<int>(2));
  for(int i=0;i<n;i++)
    cin>>xh.at(i).at(0)>>xh.at(i).at(1);
  sort(xh.begin(),xh.end());
  int64_t c=0;
  int64_t sa=0;
  queue<vector<int64_t>> q;
  for(int i=0;i<n;i++){
    int64_t x=xh.at(i).at(0);
    int64_t h=xh.at(i).at(1);
    while(!q.empty()){
      int64_t qx=q.front().at(0);
      int64_t qa=q.front().at(1);
      if(x>qx){
        sa-=qa;
        q.pop();
      }else
        break;
    }
    h=max(h-sa,(int64_t)0);
    c+=(h+a-1)/a;
    sa+=(h+a-1)/a*a;
    q.push({x+d*2,(h+a-1)/a*a});
  }
  cout<<c<<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<vector<int>> xh(n,vector<int>(2));
  for(int i=0;i<n;i++)
    cin>>xh.at(i).at(0)>>xh.at(i).at(1);
  sort(xh.begin(),xh.end());
  int64_t c=0;
  int64_t sa=0;
  queue<vector<int64_t>> q;
  for(int i=0;i<n;i++){
    int64_t x=xh.at(i).at(0);
    int64_t h=xh.at(i).at(1);
    while(!q.empty()){
      int64_t qx=q.front().at(0);
      int64_t qa=q.front().at(1);
      if(x>qx){
        sa-=qa;
        q.pop();
      }else
        break;
    }
    h=max(h-sa,(int64_t)0);
    c+=(h+a-1)/a;
    sa+=(h+a-1)/a*a;
    q.push({x+d*2,(h+a-1)/a*a});
  }
  cout<<c<<endl;
}
```