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 = 4 * 105000;
int n, can;
int a[MAXN];
int sum[MAXN];
int st[MAXN], en[MAXN];
void solve() {
  scanf("%d %d", &n, &can);
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  for (int i = 0; i <= 2 * can; i++) {
    sum[i] = 0;
    st[i] = 0;
    en[i] = 0;
  }
  for (int i = 0; i < n / 2; i++) {
    int l = a[i], r = a[n - 1 - i];
    if (l > r) swap(l, r);
    sum[l + r]++;
    st[l + 1]++;
    en[r + can]++;
  }
  int ans = n;
  int ones = 0;
  for (int i = 2; i <= 2 * can; i++) {
    ones += st[i];
    int curAns = (n / 2 - ones) * 2 + ones - sum[i];
    ans = min(ans, curAns);
    ones -= en[i];
  }
  printf("%d\n", ans);
}
int main() {
  int tests;
  scanf("%d\n", &tests);
  while (tests--) {
    solve();
  }
  return 0;
}