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;
using ll = long long;
ll power(ll x, ll y);
const ll MOD = 1000000007;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ll t;
  cin >> t;
  while (t--) {
    ll n, k;
    cin >> n >> k;
    ll arr[n];
    for (int i = 0; i < n; i++) cin >> arr[i];
    unordered_map<ll, ll> f;
    ll pre[2 * k + 1];
    memset(pre, 0, sizeof(pre));
    ;
    for (int i = 0; i < n / 2; i++) {
      ll l = arr[i], r = arr[n - i - 1];
      if (r < l) swap(l, r);
      f[l + r]++;
      pre[1 + l]++;
      pre[k + r + 1]--;
    }
    for (int i = 1; i <= 2 * k; i++) pre[i] += pre[i - 1];
    ll res = n + 1;
    for (int x = 2; x <= 2 * k; x++)
      res = min(res, pre[x] - f[x] + (n / 2 - pre[x]) * 2);
    cout << res << "\n";
  }
}
ll power(ll x, ll y) {
  ll res = 1;
  x %= MOD;
  while (y > 0) {
    if (y & 1) res = (res * x) % MOD;
    y = y >> 1, x = (x * x) % MOD;
  }
  return res;
}