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

int main() {
  int n, d, a;
  cin >> n >> d >> a;
  vector<pair<int, int64_t>> xh(n);
  for (int i = 0; i < n; i++) {
    int x, h;
    cin >> x >> h;
    xh.at(i) = make_pair(x, (h + a - 1) / a);
  }
  sort(xh.begin(), xh.end());
  queue<pair<int, int64_t>> q;
  int64_t s = 0, ans = 0;
  for (int i = 0; i < n; i++) {
    while (!q.empty() && q.front().first < xh.at(i).first - d) {
      s -= q.front().second;
      q.pop();
    }
    if (!q.empty()) xh.at(i).second -= s;
    if (xh.at(i).second <= 0) continue;
    ans += xh.at(i).second;
    q.emplace(xh.at(i).first + d, xh.at(i).second);
    s += xh.at(i).second;
  }
  cout << ans << endl;
}

### Prompt

Your task is to create 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() {
  int n, d, a;
  cin >> n >> d >> a;
  vector<pair<int, int64_t>> xh(n);
  for (int i = 0; i < n; i++) {
    int x, h;
    cin >> x >> h;
    xh.at(i) = make_pair(x, (h + a - 1) / a);
  }
  sort(xh.begin(), xh.end());
  queue<pair<int, int64_t>> q;
  int64_t s = 0, ans = 0;
  for (int i = 0; i < n; i++) {
    while (!q.empty() && q.front().first < xh.at(i).first - d) {
      s -= q.front().second;
      q.pop();
    }
    if (!q.empty()) xh.at(i).second -= s;
    if (xh.at(i).second <= 0) continue;
    ans += xh.at(i).second;
    q.emplace(xh.at(i).first + d, xh.at(i).second);
    s += xh.at(i).second;
  }
  cout << ans << endl;
}

```