A permutation of length n is an array p=[p_1,p_2,...,p_n], which contains every integer from 1 to n (inclusive) and, moreover, each number appears exactly once. For example, p=[3,1,4,2,5] is a permutation of length 5.

For a given number n (n ≥ 2), find a permutation p in which absolute difference (that is, the absolute value of difference) of any two neighboring (adjacent) elements is between 2 and 4, inclusive. Formally, find such permutation p that 2 ≤ |p_i - p_{i+1}| ≤ 4 for each i (1 ≤ i < n).

Print any such permutation for the given integer n or determine that it does not exist.

Input

The first line contains an integer t (1 ≤ t ≤ 100) — the number of test cases in the input. Then t test cases follow.

Each test case is described by a single line containing an integer n (2 ≤ n ≤ 1000).

Output

Print t lines. Print a permutation that meets the given requirements. If there are several such permutations, then print any of them. If no such permutation exists, print -1.

Example

Input


6
10
2
4
6
7
13


Output


9 6 10 8 4 7 3 1 5 2 
-1
3 1 4 2 
5 3 6 2 4 1 
5 1 3 6 2 4 7 
13 9 7 11 8 4 1 3 5 2 6 10 12 
#include <bits/stdc++.h>
using namespace std;
void solve() {
  int n;
  cin >> n;
  if (n <= 3) {
    cout << -1 << '\n';
    return;
  }
  int start = n % 4;
  vector<int> order = {(start + 3) % 4, (start + 1) % 4, start,
                       (start + 2) % 4};
  bool back = 1;
  for (int i = 0; i < 4; i++) {
    if (back) {
      int index = n - start + order[i];
      while (index > 0) {
        if (index <= n) cout << index << ' ';
        index -= 4;
      }
    } else {
      int index = order[i];
      if (index == 0) index = 4;
      while (index <= n) {
        cout << index << ' ';
        index += 4;
      }
    }
    back = !back;
  }
  cout << '\n';
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int tt;
  cin >> tt;
  while (tt--) solve();
  return 0;
}
