#include <bits/stdc++.h>
#define rep(i, n) for (int i = 0; i < (n); i++)
#define all(x) (x).begin(), (x).end()
using ll = long long;
using namespace std;

int main() {
    ll N, D, A;
    cin >> N >> D >> A;
    D = D * 2;
    vector<pair<int, int>> m(N);
    rep(i, N) cin >> m[i].first >> m[i].second;
    sort(all(m));

    ll ans = 0;
    ll dmg = 0;
    queue<pair<ll, ll>> q;
    rep(i, N) {
        while (!q.empty() && m[i].first > q.front().first) {
            dmg -= q.front().second;
            q.pop();
        }
        ll cnt = max((m[i].second - dmg + A - 1) / A ,0LL);
				ans += cnt;
        dmg += cnt * A;
        q.push({m[i].first + D, cnt * A});
    }

    cout << ans << endl;
}
### 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>
#define rep(i, n) for (int i = 0; i < (n); i++)
#define all(x) (x).begin(), (x).end()
using ll = long long;
using namespace std;

int main() {
    ll N, D, A;
    cin >> N >> D >> A;
    D = D * 2;
    vector<pair<int, int>> m(N);
    rep(i, N) cin >> m[i].first >> m[i].second;
    sort(all(m));

    ll ans = 0;
    ll dmg = 0;
    queue<pair<ll, ll>> q;
    rep(i, N) {
        while (!q.empty() && m[i].first > q.front().first) {
            dmg -= q.front().second;
            q.pop();
        }
        ll cnt = max((m[i].second - dmg + A - 1) / A ,0LL);
				ans += cnt;
        dmg += cnt * A;
        q.push({m[i].first + D, cnt * A});
    }

    cout << ans << endl;
}
```