Problem:
Score : 100 points

Problem Statement

There are two buttons, one of size A and one of size B.

When you press a button of size X, you get X coins and the size of that button decreases by 1.

You will press a button twice. Here, you can press the same button twice, or press both buttons once.

At most how many coins can you get?

Constraints

All values in input are integers.

3 \leq A, B \leq 20

Input

Input is given from Standard Input in the following format:

A B

Output

Print the maximum number of coins you can get.

Sample Input 1

5 3

Sample Output 1

9

You can get 5 + 4 = 9 coins by pressing the button of size 5 twice, and this is the maximum result.

Sample Input 2

3 4

Sample Output 2

7

Sample Input 3

6 6

Sample Output 3

12

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.*;
import java.io.*;
public class Main
{
    public static void main(String []args) throws IOException {
    	FastScanner in  = new FastScanner(System.in);
    	PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)), false);
    	solve(in, out);
    	in.close();
    	out.close();
    }
    static long gcd(long a, long b){ return (b==0) ? a : gcd(b, a%b); }
    static int gcd(int a, int b){ return (b==0) ? a : gcd(b, a%b); }
    private static int fact(int n) { int ans=1; for(int i=2;i<=n;i++) ans*=i; return ans; }
    static class FastScanner{
	BufferedReader reader;
	StringTokenizer st;
	FastScanner(InputStream stream){reader=new BufferedReader(new InputStreamReader(stream));st=null;}
	String next(){while(st == null || !st.hasMoreTokens()){try{String line = reader.readLine();if(line == null){return null;}
	     st = new StringTokenizer(line);}catch (Exception e){throw new RuntimeException();}}return st.nextToken();}
	String nextLine()  {String s=null;try{s=reader.readLine();}catch(IOException e){e.printStackTrace();}return s;}
	int    nextInt()   {return Integer.parseInt(next());}
	long   nextLong()  {return Long.parseLong(next());}
	double nextDouble(){return Double.parseDouble(next());}
	char   nextChar()  {return next().charAt(0);}
	int[]  nextIntArray(int n)         {int[] arr= new int[n];   int i=0;while(i<n){arr[i++]=nextInt();}  return arr;}
	long[] nextLongArray(int n)        {long[]arr= new long[n];  int i=0;while(i<n){arr[i++]=nextLong();} return arr;}
	int[]  nextIntArrayOneBased(int n) {int[] arr= new int[n+1]; int i=1;while(i<=n){arr[i++]=nextInt();} return arr;}
	long[] nextLongArrayOneBased(int n){long[]arr= new long[n+1];int i=1;while(i<=n){arr[i++]=nextLong();}return arr;}
	void   close(){try{reader.close();}catch(IOException e){e.printStackTrace();}}
    }
    
    /********* SOLUTION STARTS HERE ************/
    
    private static void solve(FastScanner in, PrintWriter out){
        int a = in.nextInt();
        int b = in.nextInt();
        out.println(Math.max(Math.max(a+a-1,b+b-1),a+b));
    }
    
    /************* SOLUTION ENDS HERE **********/
}