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;
long long exp(long long n, long long m, long long md) {
  long long a = 1;
  while (m > 0) {
    if (m & 1) a = (a * n % md) % md;
    m /= 2;
    n = (n * n) % md;
  }
  return a % md;
}
void printarray(int a[], int n) {
  for (int i = 0; i < n; i++) cout << a[i] << " ";
  cout << "\n";
}
void printvector(vector<int> v) {
  for (auto x : v) cout << x << " ";
  cout << "\n";
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long i, j, k, n, t, temp, arr[1009];
  cin >> t;
  while (t--) {
    cin >> n;
    if (n < 4)
      cout << "-1" << endl;
    else {
      i = n / 2;
      i -= 2;
      temp = i;
      arr[i++] = 3;
      arr[i++] = 1;
      arr[i++] = 4;
      arr[i++] = 2;
      j = 0;
      for (; i < n; i++) {
        arr[i] = 5 + 2 * j;
        j++;
      }
      j = 0;
      for (i = temp - 1; i >= 0; i--) {
        arr[i] = 6 + 2 * j;
        j++;
      }
      for (i = 0; i < n; i++) cout << arr[i] << " ";
      cout << endl;
    }
  }
  return 0;
}
