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;
long long powmod(long long base, long long exponent, long long mod) {
  long long ans = 1;
  while (exponent) {
    if (exponent & 1) ans = (ans * base) % mod;
    base = (base * base) % mod;
    exponent /= 2;
  }
  return ans;
}
long long gcd(long long a, long long b) {
  if (b == 0)
    return a;
  else
    return gcd(b, a % b);
}
const int upperlimit = 1e6 + 100;
const int mod = 1e9 + 7;
int a1[4] = {2, 4, 1, 3};
int a2[5] = {1, 3, 5, 2, 4};
int a3[6] = {1, 3, 5, 2, 4, 6};
int a4[7] = {1, 3, 5, 2, 6, 4, 7};
int main() {
  int t, n;
  scanf("%d", &t);
  while (t--) {
    scanf("%d", &n);
    if (n <= 3) {
      printf("%d\n", -1);
      continue;
    }
    int k = n - 4 - (n % 4);
    for (int i = 0; i < k; i++) printf("%d ", i - (i % 4) + (a1[i % 4]));
    if (n % 4 == 0) {
      for (int i = k; i < n; i++) printf("%d ", k + a1[i - k]);
    }
    if (n % 4 == 1) {
      for (int i = k; i < n; i++) printf("%d ", k + a2[i - k]);
    }
    if (n % 4 == 2) {
      for (int i = k; i < n; i++) printf("%d ", k + a3[i - k]);
    }
    if (n % 4 == 3) {
      for (int i = k; i < n; i++) printf("%d ", k + a4[i - k]);
    }
    printf("\n");
  }
  return 0;
}
