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

pair<long, int> x[222222];
long b[222222];
int main() {
    int n, d, a;
    cin >> n >> d >> a;
    for(int i = 0; i < n; i++)
        cin >> x[i].first >> x[i].second,
            x[i].second = (x[i].second + a - 1) / a;
    sort(x, x + n);
    long c = 0, r = 0;
    for(int i = 0; i < n; i++) {
        c += b[i];
        auto t = x[i].second + c;
        if(t > 0) {
            r += t;
            c -= t;
            int j =
                upper_bound(x, x + n, make_pair(x[i].first + d + d, (int)1e9)) -
                x;
            if(j < n)
                b[j] += t;
        }
    }
    cout << r << endl;
    return 0;
}

### Prompt

Please create a solution in CPP 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;

pair<long, int> x[222222];
long b[222222];
int main() {
    int n, d, a;
    cin >> n >> d >> a;
    for(int i = 0; i < n; i++)
        cin >> x[i].first >> x[i].second,
            x[i].second = (x[i].second + a - 1) / a;
    sort(x, x + n);
    long c = 0, r = 0;
    for(int i = 0; i < n; i++) {
        c += b[i];
        auto t = x[i].second + c;
        if(t > 0) {
            r += t;
            c -= t;
            int j =
                upper_bound(x, x + n, make_pair(x[i].first + d + d, (int)1e9)) -
                x;
            if(j < n)
                b[j] += t;
        }
    }
    cout << r << endl;
    return 0;
}

```