Being a nonconformist, Volodya is displeased with the current state of things, particularly with the order of natural numbers (natural number is positive integer number). He is determined to rearrange them. But there are too many natural numbers, so Volodya decided to start with the first n. He writes down the following sequence of numbers: firstly all odd integers from 1 to n (in ascending order), then all even integers from 1 to n (also in ascending order). Help our hero to find out which number will stand at the position number k.

Input

The only line of input contains integers n and k (1 ≤ k ≤ n ≤ 1012).

Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specifier.

Output

Print the number that will stand at the position number k after Volodya's manipulations.

Examples

Input

10 3


Output

5

Input

7 7


Output

6

Note

In the first sample Volodya's sequence will look like this: {1, 3, 5, 7, 9, 2, 4, 6, 8, 10}. The third place in the sequence is therefore occupied by the number 5.

import java.io.*;
import java.util.*;

/**
 *  
 * @author akashvermaofskt
 * Coding is love <3!
 */

public class A318 {
    public static void main(String args[]) {
        try {
            long n=nextLong();
            long k=nextLong();
            
            if(n%2==0){
                if(k<=n/2){
                    bw.write((k*2 -1)+"\n");
                }else{
                    k=k-(n/2);
                    bw.write((k*2)+"\n");
                }
            }else{
                if(k <= n/2 +1){
                    bw.write((k*2 -1)+"\n");
                }else{
                    k=k-(n/2 +1);
                    bw.write((k*2)+"\n");
                }
            }
            bw.flush();
        } catch (Exception e) {
            e.printStackTrace();
        }
    }


    //TEMPLATE
    
    //******************************************************************************
    // Fast I/O 
    static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    static BufferedWriter bw = new BufferedWriter(new OutputStreamWriter(System.out));
    static StringTokenizer st = null;
    
    static String next() {
        while (st == null || !st.hasMoreElements()) {
            try {
                st = new StringTokenizer(br.readLine());
            } catch (IOException e) {
                e.printStackTrace();
            }
        }
        return st.nextToken();
    }
    
    static int nextInt() {
        return Integer.parseInt(next());
    }
    
    static long nextLong() {
        return Long.parseLong(next());
    }

    static double nextDouble() {
        return Double.parseDouble(next());
    }
    //******************************************************************************
    // Useful user datatype
    
    static class Pair{
        public int first;
        public int second;

        public Pair(int first, int second){
            this.first = first;
            this.second = second;
            }
        @Override
        public String toString() {
        return "Pair [first=" + first + ", second=" + second + "]";
        }
    }

    static Comparator csort=new Comparator<Pair>(){
                    public int compare(Pair a, Pair b) {
                        if(a.first < b.first)
                        return -1;
                        else if (a.first > b.first)
                        return 1;
                        else
                        return 0;
                    }
                };
    //******************************************************************************
    //Permutations
    // simply prints all permutation - to see how it works
    private static void printPermutations( Comparable[] c ) {
            System.out.println( Arrays.toString( c ) );
            while ( ( c = nextPermutation( c ) ) != null ) {
                    System.out.println( Arrays.toString( c ) );
            }
    }

    // modifies c to next permutation or returns null if such permutation does not exist
    private static Comparable[] nextPermutation( final Comparable[] c ) {
            // 1. finds the largest k, that c[k] < c[k+1]
            int first = getFirst( c );
            if ( first == -1 ) return null; // no greater permutation
            // 2. find last index toSwap, that c[k] < c[toSwap]
            int toSwap = c.length - 1;
            while ( c[ first ].compareTo( c[ toSwap ] ) >= 0 )
                    --toSwap;
            // 3. swap elements with indexes first and last
            swap( c, first++, toSwap );
            // 4. reverse sequence from k+1 to n (inclusive) 
            toSwap = c.length - 1;
            while ( first < toSwap )
                    swap( c, first++, toSwap-- );
            return c;
    }

    // finds the largest k, that c[k] < c[k+1]
    // if no such k exists (there is not greater permutation), return -1
    private static int getFirst( final Comparable[] c ) {
            for ( int i = c.length - 2; i >= 0; --i )
                    if ( c[ i ].compareTo( c[ i + 1 ] ) < 0 )
                            return i;
            return -1;
    }

    // swaps two elements (with indexes i and j) in array 
    private static void swap( final Comparable[] c, final int i, final int j ) {
            final Comparable tmp = c[ i ];
            c[ i ] = c[ j ];
            c[ j ] = tmp;
    }
    //******************************************************************************
    // DSU
    //Maintain an Array A with N elements and Array size for size of each set, intialize A[i]=i and size[i]=1

    static int root(int A[],int i){
        while(A[i]!=i){
            A[i]=A[A[i]];
            i=A[i];
        }
        return i;
    }

    static boolean find(int A[],int a,int b){
        if(root(A,a)==root(A,b))return true;
        else return false;
    }

    static void union(int A[],int size[],int a,int b){
        int ra=root(A,a);
        int rb=root(A,b);

        if(ra==rb)return;

        if(size[ra]<size[rb]){
            A[ra]=A[rb];
            size[rb]+=size[ra];
        }else{
            A[rb]=A[ra];
            size[ra]+=size[rb];
        }
    }

    //**************************************************************************
    //BinarySearch
    /* binary search for 5:
                v-- lower bound
         1 2 3 4 5 5 5 6 7 9
                      ^-- upper bound
        
            binary search for 8
                        v-- lower bound
       1 2 3 4 5 5 5 6 7 9
                        ^-- upper bound
    */
    //For finding greater than 
    static int upper_bound(int A[],int key){
        int first = 0;
        int last = A.length;
        int mid;

        while (first < last) {
            mid =first+(last - first)/2; 

            if (A[mid] <= key)  
                first = mid + 1; 
            else 
                last = mid;
        }

        return first;
    }
    
    //For finding greater than equal to
    static int lower_bound(int A[],int key){
        int first = 0;
        int last = A.length;
        int mid=0;
        
        while (first < last) {
            mid = first + ((last - first) >> 1); 
            if (A[mid] < key)  
                first = mid + 1; 
            else 
                last = mid;
        }
        
        return first;
    }
    //**************************************************************************
}
