A. Even Odds

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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 %I64dspecifier.

Output

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

   Sample test(s) 
 
     input 
   
复制代码10 3
10 3

     output 
   
复制代码55

     input 
   
复制代码7 7
7 7

     output 
   
复制代码66

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.

解题说明：简单的数学题，先是排列从1到不超过n的奇数，然后排列从2到不超过n的偶数，然后找出第k个数是多少。注意此题输入需要用long long类型，输出用double，否则越界。

复制代码

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#include<iostream>
#include<map>
#include<string>
#include<algorithm>
#include<cstdio>
#include<cmath>
using namespace std;


int main()
{
	long long int n,k;
	double ans;
	scanf("%lld %lld",&n,&k);
	if(n%2==0)
	{
		if(k<=n/2)
		{
			ans=2*k-1;
		}
		else
		{
			k=k-n/2;
			ans=2*k;
		}
	}
	else
	{
		if(k<=n/2+1)
		{
			ans=2*k-1;
		}
		else
		{
			k=k-n/2-1;
			ans=2*k;
		}
	}
	printf("%.0lfn",ans);
	return 0;
}