Since Grisha behaved well last year, at New Year's Eve he was visited by Ded Moroz who brought an enormous bag of gifts with him! The bag contains n sweet candies from the good ol' bakery, each labeled from 1 to n corresponding to its tastiness. No two candies have the same tastiness.

The choice of candies has a direct effect on Grisha's happiness. One can assume that he should take the tastiest ones — but no, the holiday magic turns things upside down. It is the xor-sum of tastinesses that matters, not the ordinary sum!

A xor-sum of a sequence of integers a1, a2, ..., am is defined as the bitwise XOR of all its elements: <image>, here <image> denotes the bitwise XOR operation; more about bitwise XOR can be found [here.](https://en.wikipedia.org/wiki/Bitwise_operation#XOR)

Ded Moroz warned Grisha he has more houses to visit, so Grisha can take no more than k candies from the bag. Help Grisha determine the largest xor-sum (largest xor-sum means maximum happiness!) he can obtain.

Input

The sole string contains two integers n and k (1 ≤ k ≤ n ≤ 1018).

Output

Output one number — the largest possible xor-sum.

Examples

Input

4 3


Output

7


Input

6 6


Output

7

Note

In the first sample case, one optimal answer is 1, 2 and 4, giving the xor-sum of 7.

In the second sample case, one can, for example, take all six candies and obtain the xor-sum of 7.

import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.PrintWriter;

public class B_New_Years_Eve {
    //    buffered writer
    static PrintWriter out = new PrintWriter(System.out);
    //    buffered reader
    static BufferedReader rd = new BufferedReader(new InputStreamReader(System.in));

    public static void main(String[] args) throws Exception {
        //
        String[] ar = rd.readLine().split(" ");
        long k = Long.valueOf(ar[1]), n = Long.valueOf(ar[0]);

        //
//        long ans = 0;
        long highest = Long.highestOneBit(n);
//        while(k > 0 && highest > 0) {
//            ans += highest;
//            highest /= 2;
//            k --;
//        }
        if (k == 1) System.out.println(n);
        else {
            System.out.println(2 * highest - 1);
        }

//        System.out.println(ans);
    }
}
