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.*;
import java.math.BigDecimal;
import java.math.MathContext;
import java.util.*;

import static java.math.BigDecimal.ROUND_HALF_UP;

public class Solution {
    static MyScanner sc;
    private static PrintWriter out;
    static long M2 = 1_000_000_000L + 7;

    public static void main(String[] s) throws Exception {
        StringBuilder stringBuilder = new StringBuilder();
//        stringBuilder.append("100000 1000000000 10001 100000 ");
//        for (int i = 1; i < 100000; i++) {
//            stringBuilder.append(" 10000 100000 ");
//        }
        if (stringBuilder.length() == 0) {
            sc = new MyScanner(System.in);
        } else {
            sc = new MyScanner(new BufferedReader(new StringReader(stringBuilder.toString())));
        }


        out = new PrintWriter(new OutputStreamWriter(System.out));
        initData();
        solve();
        out.flush();
    }


    private static void initData() {

    }

    private static void solve() throws IOException {
        long n = sc.nextLong();
        long k = sc.nextLong();
        if (k == 1) {
            out.println(n);
        } else {
            long l = Long.highestOneBit(n);
            out.println( l + l - 1);
        }
    }



    private static final class DU {
        int[] par;
        int[] ct;

        private DU(int n) {
            par = new int[n];
            ct = new int[n];
            for (int i = 0; i < n; i++) {
                par[i] = i;
            }
        }

        private void union(int l1, int l2) {
            int r1 = r(l1);
            int r2 = r(l2);
            if (ct[r1] > ct[r2]) {
                par[r2] = r1;
                ct[r1] += ct[r2] + 1;
            } else {
                par[r1] = r2;
                ct[r2] += ct[r1] + 1;
            }
        }

        private int r(int l2) {
            int from = l2;
            while (l2 != par[l2]) {
                l2 = par[l2];
            }
            while (from != l2) {
                int s = par[from];
                par[from] = l2;
                from = s;
            }
            return l2;
        }
    }


    private static void solveT() throws IOException {
        int t = sc.nextInt();
        while (t-- > 0) {
            solve();
        }
    }

    private static long gcd(long l, long l1) {
        if (l > l1) return gcd(l1, l);
        if (l == 0) return l1;
        return gcd(l1 % l, l);
    }

    private static long pow(long a, long b, long m) {
        if (b == 0) return 1;
        if (b == 1) return a;
        long pp = pow(a, b / 2, m);
        pp *= pp;
        pp %= m;
        return (pp * (b % 2 == 0 ? 1 : a)) % m;
    }


    static class MyScanner {
        BufferedReader br;
        StringTokenizer st;

        MyScanner(BufferedReader br) {
            this.br = br;
        }

        public MyScanner(InputStream in) {
            this(new BufferedReader(new InputStreamReader(in)));
        }

        void findToken() {
            while (st == null || !st.hasMoreTokens()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
        }

        String next() {
            findToken();
            return st.nextToken();
        }

        Integer[] nab(int n) {
            Integer[] k = new Integer[n];
            for (int i = 0; i < n; i++) {
                k[i] = sc.fi();
            }
            return k;
        }

        int[] na(int n) {
            int[] k = new int[n];
            for (int i = 0; i < n; i++) {
                k[i] = sc.fi();
            }
            return k;
        }

        long[] nl(int n) {
            long[] k = new long[n];
            for (int i = 0; i < n; i++) {
                k[i] = sc.nextLong();
            }
            return k;
        }

        int nextInt() {
            return Integer.parseInt(next());
        }

        int fi() {
            String t = next();
            int cur = 0;
            boolean n = t.charAt(0) == '-';
            for (int a = n ? 1 : 0; a < t.length(); a++) {
                cur = cur * 10 + t.charAt(a) - '0';
            }
            return n ? -cur : cur;
        }

        long nextLong() {
            return Long.parseLong(next());
        }

        double nextDouble() {
            return Double.parseDouble(next());
        }
    }


}