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

int main() {
  long N, D, A;
  cin >> N >> D >> A;
  vector<pair<long, long>> V(N);
  for (long i = 0, x, h; cin >> x >> h; i++) V.at(i) = {x, h};
  sort(V.begin(), V.end());
  queue<pair<long, long>> Q;
  long ans = 0;
  for (long i = 0, cor = 0; i < N; i++) {
    auto [pos, hp] = V.at(i);
    while (Q.size()) {
      auto [range, dmg] = Q.front();
      if (pos > range) {
        cor -= dmg;
        Q.pop();
      } else {
        break;
      }
    }
    hp -= cor;
    if (hp <= 0) continue;
    long cnt = (hp + A - 1) / A;
    ans += cnt;
    Q.push({pos + D + D, A * cnt});
    cor += A * cnt;
  }
  cout << ans << "\n";
}
### Prompt

Please formulate 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() {
  long N, D, A;
  cin >> N >> D >> A;
  vector<pair<long, long>> V(N);
  for (long i = 0, x, h; cin >> x >> h; i++) V.at(i) = {x, h};
  sort(V.begin(), V.end());
  queue<pair<long, long>> Q;
  long ans = 0;
  for (long i = 0, cor = 0; i < N; i++) {
    auto [pos, hp] = V.at(i);
    while (Q.size()) {
      auto [range, dmg] = Q.front();
      if (pos > range) {
        cor -= dmg;
        Q.pop();
      } else {
        break;
      }
    }
    hp -= cor;
    if (hp <= 0) continue;
    long cnt = (hp + A - 1) / A;
    ans += cnt;
    Q.push({pos + D + D, A * cnt});
    cor += A * cnt;
  }
  cout << ans << "\n";
}
```