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>
int par(int k, int i) {
  if (i < k / 2) {
    return -1;
  } else {
    return 1;
  }
}
void generate(long l) {
  if (l % 4 != 0) {
    printf("NO\n");
    return;
  }
  printf("YES\n");
  long k = l / 2, arr[l];
  for (long i = 0; i < k; i++) arr[i] = 2 * i + 2;
  for (long i = k; i < l; i++) arr[i] = arr[i - k] + par(k, i - k);
  for (long i = 0; i < l; i++) printf("%d ", arr[i]);
  printf("\n");
}
int main() {
  int queries;
  long len;
  scanf("%d\n", &queries);
  for (int i = 0; i < queries; i++) {
    scanf("%ld\n", &len);
    generate(len);
  }
  return 0;
}