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.
问题链接：http://codeforces.com/problemset/problem/318/A
问题简述：输入一个数n，Volodya把小于n的奇数先列出来，再把偶数列出来，问数列里的第k个数是多少
问题分析：分n为奇数偶数两种情况模拟过程
AC通过的C++语言程序如下：
#include<iostream>
using namespace std;
int main()
{
long long n, k;
cin >> n >> k;
if (n % 2 == 0)
{
if (k <= n / 2)
{
cout << 2 * k - 1;
}
else cout << 2 * (k - n / 2);
}
else
{
if (k <= (n + 1) / 2)
{
cout << 2 * k - 1;
}
else cout << 2 * (k - (n + 1) / 2);
}
return 0;
}