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 long long M = 1e9 + 7;
long long mul(long long x, long long y) { return ((x % M) * (y % M) % M); }
long long binpow(long long x, long long y) {
  long long z = 1;
  while (y) {
    if (y & 1LL) z = mul(z, x);
    x = mul(x, x);
    y >>= 1LL;
  }
  return z;
}
long long inv(long long x) { return binpow(x, M - 2); }
long long divide(long long x, long long y) { return mul(x, inv(y)); }
const int N = 2e5 + 5;
int n, k;
int a[N];
void solve() {
  vector<int> cnt(2 * k + 2);
  vector<int> pref(2 * k + 2);
  for (int i = 0; i < n / 2; ++i) ++cnt[a[i] + a[n - i - 1]];
  for (int i = 0; i < n / 2; ++i) {
    ++pref[min(a[i], a[n - i - 1]) + 1];
    --pref[max(a[i], a[n - i - 1]) + k + 1];
  }
  for (int i = 1; i <= 2 * k + 1; ++i) pref[i] += pref[i - 1];
  int ans = 1e9;
  for (int i = 2; i <= 2 * k; ++i) {
    ans = min(ans, (pref[i] - cnt[i]) + 2 * (n / 2 - pref[i]));
  }
  cout << ans << "\n";
}
void input() {
  cin >> n >> k;
  for (int i = 0; i < n; ++i) cin >> a[i];
}
void test_case() {
  input();
  solve();
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  int q, mltcase = 1;
  if (mltcase)
    cin >> q;
  else
    q = 1;
  for (int i = 1; i <= q; ++i) test_case();
}