#include<cstdio>
#include<utility>
#include<algorithm>
#include<queue>

using namespace std;

int n, d, a, c;
pair<int, int> p[200005];
queue<pair<long long, int>> q;
long long ans;

int main() {
  scanf("%d %d %d", &n, &d, &a);
  for (int i = 0; i < n; i++) {
    scanf("%d %d", &p[i].first, &p[i].second);
  }
  sort(p, p + n);
  for (int i = 0; i < n; i++) {
    while (!q.empty() && q.front().first < p[i].first) {
      c -= q.front().second;
      q.pop();
    }
    if ((p[i].second + a - 1) / a > c) {
      q.push({p[i].first + 2 * d, (p[i].second + a - 1) / a - c});
      ans += (p[i].second + a - 1) / a - c;
      c += (p[i].second + a - 1) / a - c;
    }
  }
  printf("%lld\n", 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<cstdio>
#include<utility>
#include<algorithm>
#include<queue>

using namespace std;

int n, d, a, c;
pair<int, int> p[200005];
queue<pair<long long, int>> q;
long long ans;

int main() {
  scanf("%d %d %d", &n, &d, &a);
  for (int i = 0; i < n; i++) {
    scanf("%d %d", &p[i].first, &p[i].second);
  }
  sort(p, p + n);
  for (int i = 0; i < n; i++) {
    while (!q.empty() && q.front().first < p[i].first) {
      c -= q.front().second;
      q.pop();
    }
    if ((p[i].second + a - 1) / a > c) {
      q.push({p[i].first + 2 * d, (p[i].second + a - 1) / a - c});
      ans += (p[i].second + a - 1) / a - c;
      c += (p[i].second + a - 1) / a - c;
    }
  }
  printf("%lld\n", ans);
}

```