#include <iostream>
#include <algorithm>
using namespace std;

long n, d, a, x[200020], cumsum[200020];
pair<long, long> ps[200020];

int main() {
  cin >> n >> d >> a;
  for(int i = 0; i < n; ++i) {
    cin >> ps[i].first >> ps[i].second;
    x[i] = ps[i].first;
  }
  sort(ps, ps + n);
  sort(x, x + n);
  
  long ans = 0;
  for(int i = 0; i < n; ++i) {
    if (i > 0) cumsum[i] += cumsum[i - 1];
    ps[i].second -= cumsum[i];
    if (ps[i].second > 0) {
    int idx = upper_bound(x, x + n, ps[i].first + d * 2) - x;
    long tmp = (ps[i].second + a - 1) / a;
    cumsum[i] += tmp * a;
    cumsum[idx] -= tmp * a;
    ans += tmp;
    }
  }
  cout << ans;
}
### Prompt

Please create a solution in CPP 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 <iostream>
#include <algorithm>
using namespace std;

long n, d, a, x[200020], cumsum[200020];
pair<long, long> ps[200020];

int main() {
  cin >> n >> d >> a;
  for(int i = 0; i < n; ++i) {
    cin >> ps[i].first >> ps[i].second;
    x[i] = ps[i].first;
  }
  sort(ps, ps + n);
  sort(x, x + n);
  
  long ans = 0;
  for(int i = 0; i < n; ++i) {
    if (i > 0) cumsum[i] += cumsum[i - 1];
    ps[i].second -= cumsum[i];
    if (ps[i].second > 0) {
    int idx = upper_bound(x, x + n, ps[i].first + d * 2) - x;
    long tmp = (ps[i].second + a - 1) / a;
    cumsum[i] += tmp * a;
    cumsum[idx] -= tmp * a;
    ans += tmp;
    }
  }
  cout << ans;
}
```