#include<iostream>
#include<vector>
#include<utility>
#include<algorithm>
#include<queue>
#include<cstdio>
using namespace std;
int main() {
  int N, D, A;
  cin >> N >> D >> A;
  vector<pair<int, int>> vp(N);
  for (auto& p : vp) cin >> p.first >> p.second;
  sort(vp.begin(), vp.end());

  queue<pair<int, long long>> q;
  long long ans = 0;
  long long bias = 0;
  for (auto p : vp) {
    int x = p.first;
    int h = p.second;
    for (; q.size() && q.front().first < x;) {
      bias -= q.front().second;
      q.pop();
    }
    h -= bias;
    if (h <= 0) continue;
    int num = (h + A - 1) / A;
    int dmg = num * A;
    bias += dmg;
    ans += num;
    q.push({x + 2 * D, dmg});
  }
  cout << ans << endl;
}
### Prompt

Construct a cpp code solution to the problem outlined:

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<vector>
#include<utility>
#include<algorithm>
#include<queue>
#include<cstdio>
using namespace std;
int main() {
  int N, D, A;
  cin >> N >> D >> A;
  vector<pair<int, int>> vp(N);
  for (auto& p : vp) cin >> p.first >> p.second;
  sort(vp.begin(), vp.end());

  queue<pair<int, long long>> q;
  long long ans = 0;
  long long bias = 0;
  for (auto p : vp) {
    int x = p.first;
    int h = p.second;
    for (; q.size() && q.front().first < x;) {
      bias -= q.front().second;
      q.pop();
    }
    h -= bias;
    if (h <= 0) continue;
    int num = (h + A - 1) / A;
    int dmg = num * A;
    bias += dmg;
    ans += num;
    q.push({x + 2 * D, dmg});
  }
  cout << ans << endl;
}
```