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 long long mod = 998244353;
const int Max = 8e3 + 100;
int a[Max], b[Max];
bitset<Max> c, d;
void solve() {
  int n;
  scanf("%d", &n);
  ;
  if (n == 2 || n == 3) {
    printf("-1\n");
    return;
  }
  int beg = 1;
  while (n >= 8) {
    printf("%d %d %d %d ", beg + 1, beg + 3, beg, beg + 2);
    beg += 4;
    n -= 4;
  }
  if (n == 4) {
    printf("%d %d %d %d \n", beg + 1, beg + 3, beg, beg + 2);
    return;
  } else if (n == 5) {
    printf("%d %d %d %d %d\n", beg, beg + 2, beg + 4, beg + 1, beg + 3);
  } else if (n == 6) {
    printf("%d %d %d %d %d %d\n", beg, beg + 2, beg + 5, beg + 3, beg + 1,
           beg + 4);
  } else {
    printf("%d %d %d %d %d %d %d\n", beg, beg + 2, beg + 4, beg + 6, beg + 3,
           beg + 5, beg + 1);
  }
}
int main() {
  int t;
  scanf("%d", &t);
  ;
  while (t--) {
    solve();
  }
}
