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.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;


public class NewYearsEve {
			static class FastScanner 
			{ 
				BufferedReader br; 
				StringTokenizer st; 
		
				public FastScanner() 
				{ 
					br = new BufferedReader(new
							InputStreamReader(System.in)); 
				} 
		
				String next() 
				{ 
					while (st == null || !st.hasMoreElements()) 
					{ 
						try
						{ 
							st = new StringTokenizer(br.readLine()); 
						} 
						catch (IOException e) 
						{ 
							e.printStackTrace(); 
						} 
					} 
					return st.nextToken(); 
				} 
		
				int nextInt() 
				{ 
					return Integer.parseInt(next()); 
				} 
		
				long nextLong() 
				{ 
					return Long.parseLong(next()); 
				} 
		
				double nextDouble() 
				{ 
					return Double.parseDouble(next()); 
				} 
		
				String nextLine() 
				{ 
					String str = ""; 
					try
					{ 
						str = br.readLine(); 
					} 
					catch (IOException e) 
					{ 
						e.printStackTrace(); 
					} 
					return str; 
				} 
			} 
		
		public static void main(String[] args){
		      FastScanner f=new FastScanner();
		      long n=f.nextLong();
		      long k=f.nextLong();
		      long h=Long.highestOneBit(n);
		      long temp=h;
		      long count=0;
		      while(h>0) {
		    	  h=h/2;
		    	  count++;
		      }
		      if(k==1) {
		    	  System.out.println(n);
		    	  return;
		      }
		     
		    	  System.out.println((long)Math.pow(2,count)-(long)1);
		      
		}
		
		//BRIAN KERNINGHAM TO CHECK NUMBER OF SET BITS
			//O(LOGn)
			static int setBits(int n) {
				int count=0;
				while(n>0) {
					n=n&(n-1);
					count++;
				}
				return count;
			}
			
			//0 based indexing
			static boolean KthBitSet(int n,int k) {
				int mask=1;
				mask=mask<<=k;
				if((mask&n)!=0) return true;
				else return false;
			}

}
