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;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long int t;
  cin >> t;
  while (t--) {
    long long int n;
    cin >> n;
    if (n < 4) {
      cout << "-1" << endl;
      continue;
    }
    long long int a[4] = {2, 4, 1, 3};
    long long int ans = n / 4;
    long long int ans1 = n % 4;
    if (ans1 == 0 || ans1 == 1) {
      for (long long int i = (0); i < (ans); i++)
        for (long long int j = (0); j < (4); j++) cout << a[j] + i * 4 << " ";
      if (ans1 == 1) cout << n;
      cout << endl;
    } else {
      if (ans1 == 2) {
        long long int b[6] = {1, 4, 2, 5, 3, 6};
        for (long long int i = (0); i < (ans - 1); i++)
          for (long long int j = (0); j < (4); j++) cout << a[j] + i * 4 << " ";
        for (long long int i = (0); i < (6); i++)
          cout << b[i] + (ans - 1) * 4 << " ";
      } else {
        long long int b[7] = {1, 3, 7, 4, 6, 2, 5};
        for (long long int i = (0); i < (ans - 1); i++)
          for (long long int j = (0); j < (4); j++) cout << a[j] + i * 4 << " ";
        for (long long int i = (0); i < (7); i++)
          cout << b[i] + (ans - 1) * 4 << " ";
      }
      cout << endl;
    }
  }
  return 0;
}
