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() {
  int test;
  cin >> test;
  while ((test--) > 0) {
    int n;
    cin >> n;
    if (n <= 3) {
      cout << -1 << endl;
    } else {
      int arr[n];
      int p = n / 2;
      if (n % 2 == 0) {
        for (int i = 0; i < p; i++) {
          arr[i] = (2 * i) + 1;
        }
        int temp = arr[p - 1];
        arr[p - 1] = arr[p - 2];
        arr[p - 2] = temp;
        for (int i = p; i < (2 * p); i++) {
          arr[i] = n;
          n -= 2;
        }
        n = p * 2;
      } else {
        p = n;
        for (int i = (n / 2); i >= 0; i--) {
          arr[i] = p;
          p -= 2;
        }
        arr[(n / 2) + 1] = n - 3;
        arr[(n / 2) + 2] = n - 1;
        if (n > 5) {
          for (int i = 1; (2 * i) < (n - 3); i++) {
            arr[n - i] = 2 * i;
          }
        }
      }
      for (int i = 0; i < n; i++) {
        cout << arr[i] << " ";
      }
      cout << endl;
    }
  }
}
