#include <bits/stdc++.h>
using namespace std;
int main(){
  int N, D, A;
  cin >> N >> D >> A;
  vector<pair<int, int>> M(N);
  for (int i = 0; i < N; i++){
    int X, H;
    cin >> X >> H;
    M[i] = make_pair(X, (H + A - 1) / A);
  }
  sort(M.begin(), M.end());
  vector<int> imos(N + 1, 0);
  long long ans = 0;
  for (int i = 0; i < N; i++){
    if (i > 0){
      imos[i] += imos[i - 1];
    }
    M[i].second -= imos[i];
    if (M[i].second > 0){
      ans += M[i].second;
      auto itr = upper_bound(M.begin(), M.end(), make_pair(M[i].first + D * 2 + 1, -1));
      imos[i] += M[i].second;
      imos[itr - M.begin()] -= M[i].second;
    }
  }
  cout << ans << endl;
}
### Prompt

In Cpp, your task is to solve 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>> M(N);
  for (int i = 0; i < N; i++){
    int X, H;
    cin >> X >> H;
    M[i] = make_pair(X, (H + A - 1) / A);
  }
  sort(M.begin(), M.end());
  vector<int> imos(N + 1, 0);
  long long ans = 0;
  for (int i = 0; i < N; i++){
    if (i > 0){
      imos[i] += imos[i - 1];
    }
    M[i].second -= imos[i];
    if (M[i].second > 0){
      ans += M[i].second;
      auto itr = upper_bound(M.begin(), M.end(), make_pair(M[i].first + D * 2 + 1, -1));
      imos[i] += M[i].second;
      imos[itr - M.begin()] -= M[i].second;
    }
  }
  cout << ans << endl;
}
```