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;
int arr[200005];
int sum[100005];
int mx_pre[400005];
int mi_pre[400005];
map<long long int, int> sum_cnt;
map<long long int, int> max_cnt;
map<long long int, int> min_cnt;
void init() {
  sum_cnt.clear();
  max_cnt.clear();
  min_cnt.clear();
}
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    init();
    int n, k;
    scanf("%d%d", &n, &k);
    for (int i = 0; i < n; i++) {
      scanf("%d", &arr[i]);
    }
    for (int i = 0; i < n / 2; i++) {
      sum[i] = arr[i] + arr[n - i - 1];
      sum_cnt[sum[i]]++;
      max_cnt[max(arr[i], arr[n - i - 1])]++;
      min_cnt[min(arr[i], arr[n - i - 1])]++;
    }
    for (int i = 1; i <= 2 * k; i++) {
      mx_pre[i] = mx_pre[i - 1] + max_cnt[i];
      mi_pre[i] = mi_pre[i - 1] + min_cnt[i];
    }
    int mi = 0x3f3f3f3f;
    for (int i = 2; i <= 2 * k; i++) {
      int dif = 1;
      if (i > k) dif = i - k;
      int cnt2 = 2 * (n / 2 - mi_pre[i - 1]);
      int cnt1 = mi_pre[i - 1];
      mi = min(cnt1 + cnt2 - sum_cnt[i] + mx_pre[dif - 1], mi);
    }
    printf("%d\n", mi);
  }
  return 0;
}