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 double EPS = 1e-9;
const int INFMEM = 63;
const int INF = 1061109567;
const long long LINF = 4557430888798830399LL;
const double DINF = numeric_limits<double>::infinity();
const long long MOD = 1000000007;
const int dx[8] = {0, 0, 1, -1, 1, -1, 1, -1};
const int dy[8] = {1, -1, 0, 0, 1, -1, -1, 1};
const char dch[4] = {'R', 'L', 'D', 'U'};
const double PI = 3.141592653589793;
inline void open(string a) {
  freopen((a + ".in").c_str(), "r", stdin);
  freopen((a + ".out").c_str(), "w", stdout);
}
inline void fasterios() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
long long n;
int main() {
  int testCases;
  cin >> testCases;
  int tc = 0;
  while (++tc <= testCases) {
    cin >> n;
    if (n <= 3)
      cout << -1 << '\n';
    else {
      vector<long long> ans;
      for (int i = 1; i <= n; i += 4) {
        long long rem = n - i + 1;
        if (rem == 5) {
          ans.push_back(i);
          ans.push_back(i + 2);
          ans.push_back(i + 4);
          ans.push_back(i + 1);
          ans.push_back(i + 3);
        } else if (rem == 6) {
          ans.push_back(i);
          ans.push_back(i + 2);
          ans.push_back(i + 5);
          ans.push_back(i + 3);
          ans.push_back(i + 1);
          ans.push_back(i + 4);
        } else if (rem == 7) {
          ans.push_back(i);
          ans.push_back(i + 2);
          ans.push_back(i + 4);
          ans.push_back(i + 6);
          ans.push_back(i + 3);
          ans.push_back(i + 1);
          ans.push_back(i + 5);
        }
        if (rem != 4 && rem <= 7) break;
        ans.push_back(i + 1);
        ans.push_back(i + 3);
        ans.push_back(i);
        ans.push_back(i + 2);
      }
      for (int i = 0; i < ans.size(); i++)
        cout << ans[i] << " \n"[i == ans.size() - 1];
    }
  }
  return 0;
}
