You are given integer $n$. You have to arrange numbers from $1$ to $2n$, using each of them exactly once, on the circle, so that the following condition would be satisfied:
For every $n$ consecutive numbers on the circle write their sum on the blackboard. Then any two of written on the blackboard $2n$ numbers differ not more than by $1$.
For example, choose $n = 3$. On the left you can see an example of a valid arrangement: $1 + 4 + 5 = 10$, $4 + 5 + 2 = 11$, $5 + 2 + 3 = 10$, $2 + 3 + 6 = 11$, $3 + 6 + 1 = 10$, $6 + 1 + 4 = 11$, any two numbers differ by at most $1$. On the right you can see an invalid arrangement: for example, $5 + 1 + 6 = 12$, and $3 + 2 + 4 = 9$, $9$ and $12$ differ more than by $1$.
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-----Input-----
The first and the only line contain one integer $n$ ($1 \le n \le 10^5$).
-----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 $2n$ numbers — numbers from $1$ to $2n$ 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
3
Output
YES
1 4 5 2 3 6 
Input
4
Output
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.
n = int(input())
if n%2==0:
    print("NO")

else:
    print("YES")
    l = [0 for i in range(2*n)]
    for i in range(n):
        if i % 2 == 0:
            l[i] = i*2+1
            l[i+n] = l[i]+1
        else:
            l[i] = i*2+2
            l[i+n] = l[i]-1

    print(*l)

n = int(input())
if n%2==0:
    print("NO")

else:
    print("YES")
    l = [0 for i in range(2*n)]
    for i in range(n):
        if i % 2 == 0:
            l[i] = i*2+1
            l[i+n] = l[i]+1
        else:
            l[i] = i*2+2
            l[i+n] = l[i]-1

    print(*l)
