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.math.*;
import java.util.*;
 
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[][] XH = new long[N][2];
    for (int i = 0; i < N; i++) {
      XH[i][0] = sc.nextLong();
      XH[i][1] = sc.nextLong();
    }
    sc.close();

    Arrays.sort(XH, (a, b) -> Long.compare(a[0], b[0]));
    Queue<long[]> q = new ArrayDeque<>();
    long d2 = D * 2;
    long d = 0;
    long ans = 0;
    for (int i = 0; i < N; i++) {
      long hi = XH[i][1];
      if (hi <= 0) continue;

      long xi = XH[i][0];
      while (!q.isEmpty() && q.peek()[0] < xi) {
        d -= q.poll()[1];
      }
      hi -= d;
      XH[i][1] = Math.max(0, hi);
      if (hi <= 0) continue;

      long c = (hi + A - 1) / A;
      long ac = A * c;
      ans += c;
      hi -= ac;
      d += ac;
      q.add(new long[] {xi + d2, ac});
      XH[i][1] = Math.max(0, hi);
    }
    System.out.println(ans);
  }
}