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

Solution (in JAVA):

Solution:
import java.util.*;
import java.math.*;

public class Main{

  public static void main(String[] args) {
    Scanner sc = new Scanner(System.in);
    int n = sc.nextInt();
    long d = sc.nextLong();
    long a = sc.nextLong();
    long[][] num = new long[n][2];
    for(int i = 0; i < n; ++i) {
      num[i][0] = sc.nextLong();
      num[i][1] = sc.nextLong();
    }
    Arrays.sort(num, new Comparator<long[]>() {
      public int compare(long[] x, long[] y){
        if(x[0] > y[0]) return 1;
        else return -1;
      }
    });
    long tot = 0;
    long res = 0;
    List<long[]> list = new LinkedList<>();
    for(int i = 0; i < n; ++i) {
      while(!list.isEmpty() && num[i][0] > list.get(0)[0]) {
        tot -= list.remove(0)[1];
      }
      long cur = num[i][1] - tot;
      if(cur > 0) {
        long time = cur / a + ((cur % a == 0) ? 0 : 1);
        list.add(new long[]{num[i][0] + 2 * d, time * a});
        tot += time * a;
        res += time;
      } 
    }
    System.out.println(res);
  }
}