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 int maxn = 2e5;
const int mod = 1e9;
vector<int> graph[1000];
bool used[1000];
vector<int> ans;
void dfs(int v) {
  used[v] = 1;
  ans.push_back(v);
  for (auto to : graph[v]) {
    if (used[to]) continue;
    dfs(to);
    return;
  }
}
void solve() {
  int n;
  cin >> n;
  ans.clear();
  for (int i = 0; i < n; i++) {
    graph[i].clear();
    used[i] = false;
  }
  if (n < 4) {
    cout << -1 << endl;
    return;
  } else {
    for (int i = 0; i < n; i++) {
      if (i + 2 < n) graph[i].push_back(i + 2);
      if (i + 3 < n) graph[i].push_back(i + 3);
      if (i + 4 < n) graph[i].push_back(i + 4);
    }
    for (int i = 0; i < n; i++) {
      if (i - 2 >= 0) graph[i].push_back(i - 2);
      if (i - 3 >= 0) graph[i].push_back(i - 3);
      if (i - 4 >= 0) graph[i].push_back(i - 4);
    }
    for (int i = 0; i < n; i++) {
      dfs(i);
      if (ans.size() == n) {
        for (int j = 0; j < n; j++) cout << ans[j] + 1 << " ";
        cout << endl;
        return;
      }
      ans.clear();
      for (int j = 0; j < n; j++) used[j] = false;
    }
  }
  cout << -1 << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int q;
  cin >> q;
  while (q--) {
    solve();
  }
  return 0;
}
