#include<bits/stdc++.h>
using namespace std;
#define int long long
signed main(){
  int N,D,A;
  cin>>N>>D>>A;
  vector<pair<int,int>> B(N);
  for(pair<int,int> &p:B)
    cin>>p.first>>p.second;
  sort(B.begin(),B.end());
  vector<int> C(N+2,0);
  int ans=0;
  for(int i=0;i<N;i++){
    C[i+1]+=C[i];
    B[i].second-=C[i+1];
    if(B[i].second<=0)
      continue;
    int K=(B[i].second+A-1)/A;
    ans+=K;
    C[i+1]+=K*A;
    C[(upper_bound(B.begin(),B.end(),make_pair(B[i].first+D*2,1000000000LL))-B.begin())+1]-=K*A;
  }
  cout<<ans<<endl;
}
### Prompt

Please provide a CPP coded solution to the problem described below:

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;
#define int long long
signed main(){
  int N,D,A;
  cin>>N>>D>>A;
  vector<pair<int,int>> B(N);
  for(pair<int,int> &p:B)
    cin>>p.first>>p.second;
  sort(B.begin(),B.end());
  vector<int> C(N+2,0);
  int ans=0;
  for(int i=0;i<N;i++){
    C[i+1]+=C[i];
    B[i].second-=C[i+1];
    if(B[i].second<=0)
      continue;
    int K=(B[i].second+A-1)/A;
    ans+=K;
    C[i+1]+=K*A;
    C[(upper_bound(B.begin(),B.end(),make_pair(B[i].first+D*2,1000000000LL))-B.begin())+1]-=K*A;
  }
  cout<<ans<<endl;
}
```