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>
const int maxm = 2e5 + 100;
using namespace std;
vector<int> G[maxm];
int visited[maxm];
int dis[maxm];
long long power(long long a, long long n) {
  long long res = 1;
  while (n) {
    if (n % 2)
      res = res * a, n--;
    else
      a = a * a, n = n / 2;
  }
  return res;
}
void dfs(int n) {
  visited[n] = 1;
  for (auto i : G[n]) {
    if (visited[i] == 0) dfs(i);
  }
}
bool isPrime(long long n) {
  for (int i = 2; i <= sqrt(n); i++) {
    if (n % i == 0) return false;
  }
  return true;
}
long long powerOf2(long long n) {
  long long count = 0;
  while (n != 1) {
    count++;
    n = n / 2;
  }
  return count;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    if (n == 2 or n == 3)
      cout << "-1\n";
    else if (n == 4)
      cout << "2"
           << " 4 "
           << "1 "
           << "3\n";
    else {
      if (n % 2 == 0) {
        long long x = n;
        while (x >= 2) {
          cout << x << " ";
          x -= 2;
        }
        cout << "5 1 3 ";
        for (int i = 7; i <= n - 1; i += 2) cout << i << " ";
        cout << "\n";
      } else {
        long long x = n - 1;
        while (x >= 2) {
          cout << x << " ";
          x -= 2;
        }
        cout << "5 1 3 ";
        for (int i = 7; i <= n; i += 2) cout << i << " ";
        cout << "\n";
      }
    }
  }
  return 0;
}
