CodeForces - 318A Even Odds

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概述

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.

Examples

input

Copy
10 3

output

Copy
5

input

Copy
7 7

output

Copy
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.

题意：

从1到n的奇数，从1到n之间的偶数，排列在一起，找到第k个数

输入包含 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.

#include <iostream>
#include <cstring>
#include <string>
#include <algorithm>
#include <map>
using namespace std;
int main()
{
long long n,k;
cin>>n>>k;
if(n%2==0)//ou shu
{
if(n/2>=k)//ji shu
{
cout<<2*k-1<<endl;
}else if(n/2<k)//ou shu
{
k=k-(n/2);
cout<<2*k;
}
}else//JI SHU
{
if((n/2+1)>=k)// 奇数个数>=k
{
cout << 2*k-1<<endl;
return 0;
}else if((n/2+1)<k)
{
k=k-(n/2+1);
cout<<2*k<<endl;
}
}
return 0;
}

不能用数组。。会爆