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;
auto clk = clock();
const long long limit = 1e5 + 7;
int main() {
  long long t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    long long arr[n];
    if (n < 4) {
      cout << -1;
      cout << '\n';
      continue;
    }
    if (n == 4) {
      cout << "3 1 4 2\n";
      continue;
    }
    long long m = ((n - 1) / 2);
    arr[m] = 5;
    arr[m - 1] = 2;
    arr[m - 2] = 4;
    arr[m + 1] = 1;
    arr[m + 2] = 3;
    for (long long i = 0; i < m - 2; i++) {
      if (n % 2 == 1)
        arr[i] = n - 2 * i;
      else {
        arr[i] = n - 1 - 2 * i;
      }
    }
    long long t = 0;
    for (int i = n - 1; i > m + 2; i--) {
      if (n % 2 == 1)
        arr[i] = (n - 1) - 2 * t;
      else {
        arr[i] = (n)-2 * t;
      }
      t++;
    }
    for (long long i = 0; i < n; i++) cout << arr[i] << " ";
    cout << '\n';
  }
}
