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

int main() {
    int64_t n, d, a;
    cin >> n >> d >> a;
    vector<pair<int64_t, int64_t>> xh(n);
    for (auto& e : xh) cin >> e.first >> e.second;
    sort(begin(xh), end(xh));

    vector<int64_t> damage(n+1);
    int64_t ans = 0;
    for (int64_t l = 0, r = 0; l < n; l++) {
        while (r < n && xh[r].first - xh[l].first <= 2 * d) r++;
        int64_t tb = max(0l, xh[l].second - damage[l] + a - 1l) / a, ta = tb * a;
        ans += tb, damage[l] += ta, damage[r] -= ta, damage[l+1] += damage[l];
    }

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

int main() {
    int64_t n, d, a;
    cin >> n >> d >> a;
    vector<pair<int64_t, int64_t>> xh(n);
    for (auto& e : xh) cin >> e.first >> e.second;
    sort(begin(xh), end(xh));

    vector<int64_t> damage(n+1);
    int64_t ans = 0;
    for (int64_t l = 0, r = 0; l < n; l++) {
        while (r < n && xh[r].first - xh[l].first <= 2 * d) r++;
        int64_t tb = max(0l, xh[l].second - damage[l] + a - 1l) / a, ta = tb * a;
        ans += tb, damage[l] += ta, damage[r] -= ta, damage[l+1] += damage[l];
    }

    cout << ans << endl;
}

```