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 main() {
  long long t;
  scanf("%lld", &t);
  while (t--) {
    int n, k;
    cin >> n >> k;
    vector<long long> a(n);
    for (auto &it : a) cin >> it;
    map<int, int> cnt;
    for (long long i = 0; i < n / 2; i++) cnt[a[i] + a[n - i - 1]]++;
    int pre[2 * k + 5];
    memset(pre, 0, sizeof(pre));
    for (long long i = 0; i < n / 2; i++) {
      int l1 = 1 + a[i], r1 = k + a[i];
      int l2 = 1 + a[n - i - 1], r2 = k + a[n - i - 1];
      pre[min(l1, l2)]++;
      pre[max(r1, r2) + 1]--;
    }
    for (long long i = 1; i < 2 * k + 2; i++) pre[i] += pre[i - 1];
    int ans = 1e9;
    for (long long i = 2; i < 2 * k + 1; i++)
      ans = min(ans, (pre[i] - cnt[i]) + (n / 2 - pre[i]) * 2);
    cout << ans << endl;
  }
}