#include <iostream>
#include <vector>
#include <algorithm>
#include <queue>
using namespace std;

int n, d, a, x, h;
long long s, ans;
vector <pair<int, int> > v;
queue <pair<int, long long> > q;

int main(){
    cin >> n >> d >> a;
    for (int i = 0; i < n; i++){
        cin >> x >> h;
        v.push_back({x, (h-1)/a+1});
    }
    sort(v.begin(), v.end());
    for (int i = 0; i < n; i++){
        while (!q.empty() && q.front().second < v[i].first){
            s -= q.front().first;
            q.pop();
        }
        if (s < v[i].second){
            q.push({v[i].second-s, v[i].first+d*2});
            ans += v[i].second - s;
            s = v[i].second;
        }
    }
    cout << ans << "\n";
}

### Prompt

In CPP, your task is to solve 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 <iostream>
#include <vector>
#include <algorithm>
#include <queue>
using namespace std;

int n, d, a, x, h;
long long s, ans;
vector <pair<int, int> > v;
queue <pair<int, long long> > q;

int main(){
    cin >> n >> d >> a;
    for (int i = 0; i < n; i++){
        cin >> x >> h;
        v.push_back({x, (h-1)/a+1});
    }
    sort(v.begin(), v.end());
    for (int i = 0; i < n; i++){
        while (!q.empty() && q.front().second < v[i].first){
            s -= q.front().first;
            q.pop();
        }
        if (s < v[i].second){
            q.push({v[i].second-s, v[i].first+d*2});
            ans += v[i].second - s;
            s = v[i].second;
        }
    }
    cout << ans << "\n";
}

```