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

const int64_t INF=1e9+7;

int main(){
  int64_t N,D,A;
  cin>>N>>D>>A;
  vector<pair<int64_t,int64_t>> p(N);
  for(int i=0;i<N;++i){
    cin>>p[i].first>>p[i].second;
  }
  sort(p.begin(),p.end());

  vector<int64_t> dam(N+1,0);
  int64_t ans=0;

  for(int i=0;i<N;++i){
    if(dam[i]<p[i].second){
    int j=upper_bound(p.begin(),p.end(),make_pair(p[i].first+2*D,INF))-p.begin();
    int64_t need=(p[i].second-dam[i]+(A-1))/A;
    dam[i] += need*A;
    dam[j] -= need*A;
    ans += need;
  }
  dam[i+1] += dam[i];
  }

  cout<<ans<<endl;
}
### Prompt

Construct a CPP code solution to the problem outlined:

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;

const int64_t INF=1e9+7;

int main(){
  int64_t N,D,A;
  cin>>N>>D>>A;
  vector<pair<int64_t,int64_t>> p(N);
  for(int i=0;i<N;++i){
    cin>>p[i].first>>p[i].second;
  }
  sort(p.begin(),p.end());

  vector<int64_t> dam(N+1,0);
  int64_t ans=0;

  for(int i=0;i<N;++i){
    if(dam[i]<p[i].second){
    int j=upper_bound(p.begin(),p.end(),make_pair(p[i].first+2*D,INF))-p.begin();
    int64_t need=(p[i].second-dam[i]+(A-1))/A;
    dam[i] += need*A;
    dam[j] -= need*A;
    ans += need;
  }
  dam[i+1] += dam[i];
  }

  cout<<ans<<endl;
}
```