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;
void solve() {
  long long n;
  cin >> n;
  if (n <= 3) {
    cout << -1 << endl;
    return;
  }
  deque<int> dq;
  dq.push_front(3);
  dq.push_front(1);
  dq.push_front(4);
  dq.push_front(2);
  long long sum = 10;
  long long b = 1;
  while (dq.size() < n) {
    if (b == 1) {
      dq.push_back(dq.back() + 2);
      sum += dq.back();
      b = 0;
    } else {
      long long e = dq.size() + 1;
      long long rsum = ((e) * (e + 1)) / 2;
      dq.push_front(rsum - sum);
      sum = rsum;
      b = 1;
    }
  }
  while (!dq.empty()) {
    cout << dq.front() << " ";
    dq.pop_front();
  }
  cout << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t;
  cin >> t;
  while (t) {
    solve();
    t--;
  }
  return 0;
}
