Problem: You are given a positive integer n, it is guaranteed that n is even (i.e. divisible by 2).

You want to construct the array a of length n such that: 

  * The first n/2 elements of a are even (divisible by 2); 
  * the second n/2 elements of a are odd (not divisible by 2); 
  * all elements of a are distinct and positive; 
  * the sum of the first half equals to the sum of the second half (∑_{i=1}^{n/2} a_i = ∑_{i=n/2 + 1}^{n} a_i). 



If there are multiple answers, you can print any. It is not guaranteed that the answer exists.

You have to answer t independent test cases.

Input

The first line of the input contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Then t test cases follow.

The only line of the test case contains one integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the length of the array. It is guaranteed that that n is even (i.e. divisible by 2).

It is guaranteed that the sum of n over all test cases does not exceed 2 ⋅ 10^5 (∑ n ≤ 2 ⋅ 10^5).

Output

For each test case, print the answer — "NO" (without quotes), if there is no suitable answer for the given test case or "YES" in the first line and any suitable array a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9) satisfying conditions from the problem statement on the second line.

Example

Input


5
2
4
6
8
10


Output


NO
YES
2 4 1 5
NO
YES
2 4 6 8 1 3 5 11
NO

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 2e5 + 10;
int a[MAXN];
const int MAXM = 4e5 + 10;
int ans[MAXM];
void add(int x, int y, int d) {
  ans[x] += d;
  ans[y + 1] -= d;
}
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    memset(ans, 0, sizeof(ans));
    int n, i, k;
    scanf("%d %d", &n, &k);
    for (i = 1; i <= n; i++) {
      scanf("%d", &a[i]);
    }
    for (i = 1; i <= n / 2; i++) {
      int x = a[i];
      int y = a[n - i + 1];
      if (x > y) swap(x, y);
      add(0, x, 2);
      add(x + 1, x + y - 1, 1);
      add(x + y + 1, y + k, 1);
      add(y + k + 1, 2 * k, 2);
    }
    int res = ans[0];
    for (i = 1; i <= 2 * k; i++) {
      ans[i] += ans[i - 1];
      res = min(res, ans[i]);
    }
    printf("%d\n", res);
  }
  return 0;
}