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

#define int long long int

signed main() {
    int n, d, a;
    cin >> n >> d >> a;
    vector<vector<int>> v;
    vector<int> c(n + 2, 0);
    for(int i = 0; i < n; i++) {
        int x, y;
        cin >> x >> y;
        v.push_back({ x, y });
    }
    sort(v.begin(), v.end());
    int j = 0, ans = 0;
    for(int i = 0; i < n; i++) {
        while(j < n && v[j][0] - v[i][0] <= 2 * d)
            j++;
        int need = (v[i][1] - c[i] * a + a - 1) / (a);
        need = max(need, 0ll);
        ans += need;
        c[i] += need;
        c[j] -= need;
        c[i + 1] += c[i];
    }
    cout << ans;
}
### Prompt

Create a solution in CPP for 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;

#define int long long int

signed main() {
    int n, d, a;
    cin >> n >> d >> a;
    vector<vector<int>> v;
    vector<int> c(n + 2, 0);
    for(int i = 0; i < n; i++) {
        int x, y;
        cin >> x >> y;
        v.push_back({ x, y });
    }
    sort(v.begin(), v.end());
    int j = 0, ans = 0;
    for(int i = 0; i < n; i++) {
        while(j < n && v[j][0] - v[i][0] <= 2 * d)
            j++;
        int need = (v[i][1] - c[i] * a + a - 1) / (a);
        need = max(need, 0ll);
        ans += need;
        c[i] += need;
        c[j] -= need;
        c[i + 1] += c[i];
    }
    cout << ans;
}
```