#include <bits/stdc++.h>
#define P pair<long long, long long>

using namespace std;

int main(){
  long long n, d, a;
  P p[200009];
  cin >> n >> d >> a;
  for(int i=0;i<n;i++){
    long long x, h;
    cin >> x >> h;
    p[i] = P(x, h);
  }
  sort(p, p+n);
  priority_queue<P, vector<P>, greater<P> > q;
  long long total=0;
  long long cnt=0;
  for(int i=0;i<n;i++){
    while(!q.empty()&&q.top().first<p[i].first){
      total -= q.top().second;q.pop();
    }
    if(total<p[i].second){
      long long add = (p[i].second-total-1)/a + 1;
      cnt += add;
      total += add*a;
      q.push(P(p[i].first+2*d, add*a));
    }
  }
  cout << cnt << endl;
}

### 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>
#define P pair<long long, long long>

using namespace std;

int main(){
  long long n, d, a;
  P p[200009];
  cin >> n >> d >> a;
  for(int i=0;i<n;i++){
    long long x, h;
    cin >> x >> h;
    p[i] = P(x, h);
  }
  sort(p, p+n);
  priority_queue<P, vector<P>, greater<P> > q;
  long long total=0;
  long long cnt=0;
  for(int i=0;i<n;i++){
    while(!q.empty()&&q.top().first<p[i].first){
      total -= q.top().second;q.pop();
    }
    if(total<p[i].second){
      long long add = (p[i].second-total-1)/a + 1;
      cnt += add;
      total += add*a;
      q.push(P(p[i].first+2*d, add*a));
    }
  }
  cout << cnt << endl;
}

```