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:
/*
    Those who cannot remember the past are condemned to repeat it.
                                            - Dynamic Programming
    
    I never lose, I either win or I learn.
                - Competitive Programming

*/

import javax.print.DocFlavor;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.StringTokenizer;

class Pair {
    long x;
    long y;
    Pair(long u, long v) {
        x = u;
        y = v;
    }
}

public class Main {
    public static void main(String[] args) throws Exception {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st;
        PrintWriter pw = new PrintWriter(System.out);

        st = new StringTokenizer(br.readLine());
        int n = Integer.parseInt(st.nextToken());
        long d = Long.parseLong(st.nextToken());
        long a = Long.parseLong(st.nextToken());

        Pair[] p = new Pair[n];

        for(int i=0 ; i<n ; i++) {
            st = new StringTokenizer(br.readLine());
            long u = Long.parseLong(st.nextToken());
            long v = Long.parseLong(st.nextToken());
            p[i] = new Pair(u, v);
        }

        Arrays.sort(p, (o1, o2) -> (int)(o1.x-o2.x));

        int l = 0;
        int r = 0;
        long ans = 0;
        long reduce = 0;
        long[] lazy = new long[n+2];
        for(int i=0 ; i<n ; i++) {
            reduce += lazy[i];
            p[i].y -= reduce*a;
            if(p[i].y <= 0) continue;
            long val = p[i].x;
            long go_to = val+2*d;
            while(r < n && p[r].x <= go_to) {
                r++;
            }
            long reduce_at_this_step = p[i].y%a == 0 ? p[i].y/a : p[i].y/a+1;
            lazy[r] += -reduce_at_this_step;
            reduce += reduce_at_this_step;
            ans += reduce_at_this_step;
        }

        pw.println(ans);
        
        pw.flush();
        pw.close();
    }
}