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 = 2000;
const long long inf = 1e18;
bool vis[maxn];
bool dp[2000][2000];
int ans[maxn], n;
bool solve(int cur, int cnt) {
  if (cnt == n) return true;
  if (dp[cur][cnt]) return false;
  dp[cur][cnt] = true;
  for (int i = 1; i <= 4; i++) {
    if (vis[cur + i] == false and i >= 2 and (cur + i) <= n) {
      ans[cnt] = i + cur;
      vis[cur + i] = true;
      if (solve(i + cur, cnt + 1)) return true;
      vis[cur + i] = false;
    }
  }
  for (int i = 1; i <= 4; i++) {
    if ((cur - i) >= 1 and vis[cur - i] == false and i >= 2) {
      ans[cnt] = cur - i;
      vis[cur - i] = true;
      if (solve(cur - i, cnt + 1)) return true;
      vis[cur - i] = false;
    }
  }
  return false;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  int test;
  cin >> test;
  while (test--) {
    cin >> n;
    bool fl = false;
    for (int i = 1; i <= n; i++) {
      ans[0] = i;
      memset(vis, false, sizeof vis);
      memset(dp, false, sizeof dp);
      vis[i] = true;
      if (solve(i, 1)) {
        for (int i = 0; i < n; i++) {
          cout << ans[i] << " ";
        }
        fl = true;
        break;
      }
    }
    if (!fl) cout << -1;
    cout << endl;
    memset(dp, false, sizeof dp);
  }
}
