A. Almost Equal

A. Almost Equal

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given integer nn. You have to arrange numbers from 11 to 2n2n, using each of them exactly once, on the circle, so that the following condition would be satisfied:

For every nn consecutive numbers on the circle write their sum on the blackboard. Then any two of written on the blackboard 2n2n numbers differ not more than by 11.

For example, choose n=3n=3. On the left you can see an example of a valid arrangement: 1+4+5=101+4+5=10, 4+5+2=114+5+2=11, 5+2+3=105+2+3=10, 2+3+6=112+3+6=11, 3+6+1=103+6+1=10, 6+1+4=116+1+4=11, any two numbers differ by at most 11. On the right you can see an invalid arrangement: for example, 5+1+6=125+1+6=12, and 3+2+4=93+2+4=9, 99 and 1212 differ more than by 11.

Input

The first and the only line contain one integer nn (1≤n≤1051≤n≤105).

Output

If there is no solution, output "NO" in the first line.

If there is a solution, output "YES" in the first line. In the second line output 2n2n numbers — numbers from 11 to 2n2n in the order they will stay in the circle. Each number should appear only once. If there are several solutions, you can output any of them.

Examples

input

Copy

3

output

Copy

YES
1 4 5 2 3 6 

input

Copy

4

output

Copy

NO

Note

Example from the statement is shown for the first example.

It can be proved that there is no solution in the second example.

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这种规律题目一定是线性推的，也就是从答案1一直推到答案2n。不难发现，1正对着2,3正对着4，5正对着6，并且发现这在坐标上就是+n的位置差别。

且是1 2 4 3 5 6

一大一小排列

# include
# include
using namespace std;
 
int ans[100000*2+10];
 
int main()
{
 
    int n;
 
    cin>>n;
 
    if(n%2)
    {
        cout<<"YES"<
 
        int now=1;
 
        for(int i=1;i<=n;i++)
        {
            if(i%2)
            {
                ans[i]=now;
                ans[i+n]=now+1;
 
                now+=2;
            }
            else
            {
                ans[i]=now+1;
                ans[i+n]=now;
                now+=2;
            }
        }
 
        for(int i=1;i<=2*n;i++)
        {
            cout<" ";
        }
 
 
    }
    else
    {
        cout<<"NO"<
    }
 
 
    return 0;
}