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 ≤ n ≤ 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.
#include <bits/stdc++.h>
using namespace std;
const int prime = 999983;
const int INF = 0x7FFFFFFF;
const long long INFF = 0x7FFFFFFFFFFFFFFF;
const double pi = acos(-1.0);
const double inf = 1e18;
const double eps = 1e-6;
const long long mod = 1e9 + 7;
long long qpow(long long a, long long b) {
  long long s = 1;
  while (b > 0) {
    if (b & 1) s = s * a % mod;
    a = a * a % mod;
    b >>= 1;
  }
  return s;
}
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
int dr[2][4] = {1, -1, 0, 0, 0, 0, -1, 1};
const int maxn = 2e5 + 10;
int a[maxn];
int main(void) {
  int n;
  cin >> n;
  if (n & 1) {
    cout << "YES" << endl;
    a[1] = 1;
    for (int i = 2; i <= n; i += 2) {
      a[i] = a[i - 1] + 3;
      a[i + 1] = a[i] + 1;
    }
    a[n + 1] = 2;
    a[n + 2] = 3;
    for (int i = n + 3; i <= 2 * n; i += 2) {
      a[i] = a[i - 1] + 3;
      a[i + 1] = a[i] + 1;
    }
    a[2 * n] = 2 * n;
    for (int i = 1; i <= 2 * n; ++i) printf("%d ", a[i]);
  } else
    puts("NO");
  return 0;
}
