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;
void solve() {
  long long n;
  cin >> n;
  long long a[n + 1], i1;
  for (long long i = 1; i <= n; i++) a[i] = i;
  if (n < 4) {
    cout << -1 << "\n";
    return;
  }
  if (n % 4 == 0) {
    i1 = 1;
  } else if (n % 4 == 1) {
    a[1] = 2;
    a[2] = 4;
    a[3] = 1;
    a[4] = 3;
    i1 = 6;
  } else if (n % 4 == 2) {
    a[2] = 3;
    a[3] = 5;
    a[4] = 2;
    a[5] = 4;
    i1 = 7;
  } else {
    a[1] = 2;
    a[2] = 4;
    a[3] = 1;
    a[4] = 5;
    a[5] = 7;
    a[6] = 3;
    a[7] = 6;
    i1 = 8;
  }
  long long i2 = i1 + 1, i3 = i2 + 1, i4 = i3 + 1;
  while (i4 <= n) {
    long long p = a[i1], q = a[i2], r = a[i3], s = a[i4];
    a[i1] = q;
    a[i2] = s;
    a[i3] = p;
    a[i4] = r;
    i1 += 4;
    i2 += 4;
    i3 += 4;
    i4 += 4;
  }
  for (long long i = 1; i <= n; i++) cout << a[i] << " ";
  cout << "\n";
}
int main() {
  ios_base ::sync_with_stdio(false);
  cin.tie(NULL);
  int t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}
