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 printv(vector<long long> v) {
  for (long long i = 0; i < v.size(); i++) cout << v[i] << ' ';
  cout << '\n';
}
long long bem(long long m, long long n, long long mod) {
  long long res = 1;
  while (n) {
    if (n % 2 != 0) {
      res *= m;
      res %= mod;
    }
    m *= m;
    m %= mod;
    n /= 2;
  }
  return res;
}
long long be(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b % 2 != 0) res *= a;
    a *= a;
    b /= 2;
  }
  return res;
}
void sieve(long long n, vector<bool> &prime) {
  prime[0] = false;
  prime[1] = false;
  for (long long i = 2; i * i <= n; i++) {
    if (prime[i] == true) {
      for (long long j = i * i; j <= n; j += i) prime[j] = false;
    }
  }
}
void solve() {
  long long n;
  cin >> n;
  if (n % 4 != 0) {
    cout << "NO\n";
  } else {
    cout << "YES\n";
    long long z = n / 4;
    for (long long i = (long long)0; i <= (long long)z - 1; i++) {
      cout << 10 * i + 2 << ' ' << 10 * i + 4 << ' ';
    }
    for (long long i = (long long)0; i <= (long long)z - 1; i++) {
      cout << 10 * i + 1 << ' ' << 10 * i + 5 << ' ';
    }
    cout << '\n';
  }
}
int32_t main() {
  long long t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
}