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;
const int N = 1 * 1000 + 1;
int ans[N];
int main() {
  int T;
  cin >> T;
  while (T--) {
    int n;
    cin >> n;
    memset(ans, 0, sizeof(ans));
    if (n < 4) {
      cout << -1 << endl;
      continue;
    }
    ans[1] = 2, ans[2] = 4, ans[3] = 1, ans[4] = 3;
    if (n % 4 == 0)
      for (int i = 5; i <= n; i++) ans[i] = ans[i - 4] + 4;
    if (n % 4 == 1) {
      for (int i = 5; i < n; i++) ans[i] = ans[i - 4] + 4;
      ans[n] = n;
    }
    if (n % 4 == 2) {
      for (int i = 5; i < n - 2; i++) ans[i] = ans[i - 4] + 4;
      ans[n - 2] = n - 1, ans[n - 1] = n - 3, ans[n] = n;
    }
    if (n % 4 == 3) {
      for (int i = 5; i < n - 3; i++) ans[i] = ans[i - 4] + 4;
      ans[n - 3] = n - 2, ans[n - 2] = n, ans[n - 1] = n - 4, ans[n] = n - 1;
    }
    for (int i = 1; i <= n; i++) cout << ans[i] << ' ';
    cout << endl;
  }
}
