#include <bits/stdc++.h>
using namespace std;
typedef pair<long long, long long> P;

int main(void) {
  long long N, D, A;
  cin >> N >> D >> A;
  map<P, long long> m;
  for (int i = 0; i < N; i++) {
    long long X, H;
    cin >> X >> H;
    m[P(X, 2)] = H;
  }
  long long ans = 0;
  long long current = 0;
  for (auto e : m) {
    if (e.first.second == 1) {
      current += e.second;
    } else {
      if (e.second <= current) continue;
      long long cnt = (e.second - current + A - 1) / A;
      ans += cnt;
      m[P(e.first.first + 2 * D + 1, 1)] = -cnt * A;
      current += cnt * A;
    }
  }

  cout << ans << endl;
  return 0;
}
### 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;
typedef pair<long long, long long> P;

int main(void) {
  long long N, D, A;
  cin >> N >> D >> A;
  map<P, long long> m;
  for (int i = 0; i < N; i++) {
    long long X, H;
    cin >> X >> H;
    m[P(X, 2)] = H;
  }
  long long ans = 0;
  long long current = 0;
  for (auto e : m) {
    if (e.first.second == 1) {
      current += e.second;
    } else {
      if (e.second <= current) continue;
      long long cnt = (e.second - current + A - 1) / A;
      ans += cnt;
      m[P(e.first.first + 2 * D + 1, 1)] = -cnt * A;
      current += cnt * A;
    }
  }

  cout << ans << endl;
  return 0;
}
```