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;
int n, t;
vector<int> v;
set<int> s;
void solve() {
  cin >> n;
  v.clear();
  s.clear();
  if (n == 4) {
    cout << "3 1 4 2" << '\n';
    return;
  }
  if (n == 5) {
    cout << "5 3 1 4 2" << '\n';
    return;
  }
  if (n < 4) {
    cout << -1 << '\n';
    return;
  } else {
    int mxev = n / 2 * 2;
    int mxev2 = mxev - 2;
    int mxev3 = mxev2 - 2;
    for (int i = 1; i <= n; i += 2) {
      v.push_back(i);
      s.insert(i);
    }
    if (n % 2) {
      v.push_back(mxev2);
      v.push_back(mxev);
      s.insert(mxev2);
      s.insert(mxev);
      for (int i = n - 1; i >= 1; i -= 2)
        if (s.find(i) == s.end()) {
          s.insert(i);
          v.push_back(i);
        }
    } else {
      v.push_back(mxev3);
      v.push_back(mxev);
      s.insert(mxev3);
      s.insert(mxev);
      for (int i = n; i >= 1; i -= 2)
        if (s.find(i) == s.end()) {
          s.insert(i);
          v.push_back(i);
        }
    }
    for (int u : v) cout << u << ' ';
    cout << '\n';
  }
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> t;
  while (t--) {
    solve();
  }
}
