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;
inline int read() {
  int sum = 0, ff = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') ff = -1;
    ch = getchar();
  }
  while (isdigit(ch)) sum = sum * 10 + (ch ^ 48), ch = getchar();
  return sum * ff;
}
int Q, n, a[10005], cnt;
int main() {
  Q = read();
  for (; Q--;) {
    n = read();
    cnt = 0;
    if (n == 4) {
      puts("3 1 4 2");
      continue;
    }
    if (n <= 3)
      printf("-1");
    else {
      if (n & 1) {
        for (int j = n; j >= 1; j--)
          if (j & 1) a[++cnt] = j;
        a[++cnt] = 4, a[++cnt] = 2;
        if (n >= 6)
          for (int j = 6; j <= n; j++)
            if (j % 2 == 0) a[++cnt] = j;
      } else {
        for (int j = n; j >= 1; j--)
          if (j % 2 == 0) a[++cnt] = j;
        a[++cnt] = 5, a[++cnt] = 1, a[++cnt] = 3;
        if (n >= 7)
          for (int j = 7; j <= n; j++)
            if (j & 1) a[++cnt] = j;
      }
      for (int i = 1; i <= cnt; i++) printf("%d ", a[i]);
    }
    puts("");
  }
}
