#include <stdio.h>
#include <algorithm>
#include <utility>
using namespace std;
typedef long long ll;

int main(void) {
  ll i, j, k, n, d, ans = 0, a;
  scanf("%lld%lld%lld", &n, &d, &a);
  d *= 2;
  pair<ll, ll> m[n], p;
  ll sum[n + 1], now = 0;
  for(i = 0; i < n; ++i) scanf("%lld%lld", &m[i].first, &m[i].second), m[i].second = (m[i].second + a - 1) / a;
  sort(m, m + n);
  for(i = 0; i <= n; ++i) sum[i] = 0;
  for(i = 0; i < n; ++i) {
    now += sum[i];
    if(m[i].second > now) {
      ans += m[i].second - now;
      p = make_pair(m[i].first + d, 1e9);
      sum[upper_bound(m, m + n, p) - m] -= m[i].second - now;
      now = m[i].second;
    }
  }
  printf("%lld", ans);
  return 0;
}
### Prompt

Develop a solution in cpp to the problem described below:

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 <stdio.h>
#include <algorithm>
#include <utility>
using namespace std;
typedef long long ll;

int main(void) {
  ll i, j, k, n, d, ans = 0, a;
  scanf("%lld%lld%lld", &n, &d, &a);
  d *= 2;
  pair<ll, ll> m[n], p;
  ll sum[n + 1], now = 0;
  for(i = 0; i < n; ++i) scanf("%lld%lld", &m[i].first, &m[i].second), m[i].second = (m[i].second + a - 1) / a;
  sort(m, m + n);
  for(i = 0; i <= n; ++i) sum[i] = 0;
  for(i = 0; i < n; ++i) {
    now += sum[i];
    if(m[i].second > now) {
      ans += m[i].second - now;
      p = make_pair(m[i].first + d, 1e9);
      sum[upper_bound(m, m + n, p) - m] -= m[i].second - now;
      now = m[i].second;
    }
  }
  printf("%lld", ans);
  return 0;
}
```