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 a[1000];
void solve() {
  int z;
  cin >> z;
  if (z < 4) {
    cout << "-1\n";
    return;
  }
  for (int i = 0; i < z / 5; ++i) {
    int j = i * 5;
    a[j] = j + 3;
    a[j + 1] = j + 1;
    a[j + 2] = j + 5;
    a[j + 3] = j + 2;
    a[j + 4] = j + 4;
  }
  if (z % 5 == 1) {
    a[z - 1] = z;
  } else if (z % 5 == 2) {
    a[z - 3] = z - 1;
    a[z - 2] = z - 3;
    a[z - 1] = z;
  } else if (z % 5 == 3) {
    a[z - 4] = z - 2;
    a[z - 3] = z;
    a[z - 2] = z - 4;
    a[z - 1] = z - 1;
  } else if (z % 5 == 4) {
    a[z - 4] = z - 1;
    a[z - 3] = z - 3;
    a[z - 2] = z;
    a[z - 1] = z - 2;
  }
  for (int i = 0; i < z; ++i) cout << a[i] << " ";
  cout << "\n";
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int f;
  cin >> f;
  while (f--) solve();
}
