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.*;

public class Main{

    Scanner sc = new Scanner(System.in);

    public static void main(String[] args){
        new Main().run();
    }

    void run() {
        int n = sc.nextInt();
        int d = sc.nextInt();
        int a = sc.nextInt();
        int[][] arr = new int[n][2];

        for (int i=0; i<n; i++) {
            arr[i][0] = sc.nextInt();
            arr[i][1] = sc.nextInt();
        }

        Arrays.sort(arr, (x, y) -> x[0] - y[0]);

//        System.out.println(arr[2][0]);

        Queue<Pair> queue = new ArrayDeque<>();

        int i = 0;
        long ans = 0;
        long accD = 0;

        while (i < n) {
            while (!queue.isEmpty()) {
                Pair p = queue.peek();
                if (p.x >= arr[i][0]) break;
                accD -= p.damage;
                queue.poll();
            }

            arr[i][1] -= accD;
//            System.out.println(accD);

            if (arr[i][1] > 0) {
                long attack = (long)(arr[i][1] + a - 1) / (long) a;
                ans += attack;
                queue.add(new Pair((arr[i][0] + 2 * d), attack * a));
                accD += attack * a;
            }
            i++;
//            System.out.println(ans);
//            System.out.println(accD);
        }

        System.out.println(ans);


    }

    class Pair {
        long x;
        long damage;
        public Pair(long x, long d) {
            this.x = x;
            this.damage = d;
        }
    }

}