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 + 7;
long long t, n, k, a[maxn], v[maxn << 1], k2, res;
void add(long long x, long long t) {
  for (; x <= k2; x += x & -x) v[x] += t;
  return;
}
long long query(long long x) {
  long long res = 0;
  for (; x > 0; x ^= x & -x) res += v[x];
  return res;
}
int main() {
  scanf("%lld", &t);
  while (t--) {
    scanf("%lld%lld", &n, &k);
    k2 = k + k;
    for (int i = 1; i <= n; i++) scanf("%lld", &a[i]);
    for (int i = 2; i <= k2; i++) v[i] = 0;
    for (int i = 1; i + i <= n; i++) {
      add(2, 2);
      add(min(a[i], a[n - i + 1]) + 1, -1);
      add(a[i] + a[n - i + 1], -1);
      add(a[i] + a[n - i + 1] + 1, 1);
      add(max(a[i], a[n - i + 1]) + k + 1, 1);
    }
    res = query(2);
    for (int i = 3; i <= k2; i++) res = min(res, query(i));
    printf("%lld\n", res);
  }
  return 0;
}