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;
void files() {
  freopen("1.in", "r", stdin);
  freopen("1.out", "w", stdout);
  return;
}
const int MXN = 1e5 + 5;
const long long MOD = 1e9 + 7;
void sol() {
  int n;
  int k;
  scanf("%d %d", &n, &k);
  int x[n / 2], y[n / 2];
  for (int i = 0; i < n / 2; ++i) scanf("%d", &x[i]);
  for (int i = 0; i < n / 2; ++i) scanf("%d", &y[i]);
  map<int, int> cnt;
  for (int i = 0; i < n / 2; ++i) {
    cnt[x[i] + y[n / 2 - 1 - i]]++;
    int mx = max(x[i], y[n / 2 - 1 - i]);
    int mn = min(x[i], y[n / 2 - 1 - i]);
    x[i] = mx;
    y[n / 2 - 1 - i] = mn;
  }
  sort(x, x + n / 2);
  sort(y, y + n / 2);
  int ans = INT_MAX;
  for (int i = 2; i <= 2 * k; ++i) {
    int cur = 0;
    int idx = n / 2 - (lower_bound(x, x + n / 2, i) - x);
    int idy = n / 2 - (lower_bound(y, y + n / 2, i) - y);
    int idz = (lower_bound(x, x + n / 2, i - k) - x);
    cur += idx + idy + 2 * idz + (n / 2 - idx - idz) - cnt[i];
    ans = min(ans, cur);
  }
  printf("%d\n", ans);
}
int main() {
  int tcase = 1;
  scanf("%d", &tcase);
  while (tcase--) {
    sol();
  }
}