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>
const long long INF = 1e18;
void swap(long long &a, long long &b) {
  long long temp;
  temp = a;
  a = b;
  b = temp;
}
using namespace std;
void solve() {
  long long n, even[1000], ne;
  cin >> n;
  if (n == 2 || n == 3) {
    cout << "-1";
    return;
  }
  if (n == 4) {
    cout << "2 4 1 3";
    return;
  }
  ne = n / 2;
  for (int i = 0, j = 2; i < ne / 2; i++, j += 4) {
    even[i] = j;
    even[ne - 1 - i] = j + 2;
  }
  if (ne % 2 != 0) even[ne / 2] = ne * 2;
  for (int i = 1; i <= n; i += 2) {
    cout << i << " ";
    if (i == 5) {
      for (int j = 0; j < ne; j++) cout << even[j] << " ";
    }
  }
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    solve();
    cout << "\n";
  }
  return 0;
}
