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

int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n;
  long long d, a;
  cin >> n >> d >> a;
  vector< pair<int, long long> > v(n);
  for (int i = 0; i < n; i++) {
    cin >> v[i].first >> v[i].second;
  }
  sort(v.begin(), v.end());
  int i = 0, j = 0;
  long long ans = 0;
  vector<long long> cnt(n + 1, 0);
  for (; i < n; i++) {
    while (j < n && v[j].first - v[i].first <= 2 * d) {
      j++;
    }
    j--;
    int tmp = max(0ll, (v[i].second - cnt[i] * a + a - 1) / a);
    ans += tmp;
    cnt[i] += tmp;
    cnt[i + 1] += cnt[i];
    cnt[j + 1] -= tmp;
  }
  cout << ans << '\n';
  return 0;
} 
### Prompt

Generate 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() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n;
  long long d, a;
  cin >> n >> d >> a;
  vector< pair<int, long long> > v(n);
  for (int i = 0; i < n; i++) {
    cin >> v[i].first >> v[i].second;
  }
  sort(v.begin(), v.end());
  int i = 0, j = 0;
  long long ans = 0;
  vector<long long> cnt(n + 1, 0);
  for (; i < n; i++) {
    while (j < n && v[j].first - v[i].first <= 2 * d) {
      j++;
    }
    j--;
    int tmp = max(0ll, (v[i].second - cnt[i] * a + a - 1) / a);
    ans += tmp;
    cnt[i] += tmp;
    cnt[i + 1] += cnt[i];
    cnt[j + 1] -= tmp;
  }
  cout << ans << '\n';
  return 0;
} 
```