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;
long long int modpow(long long int a, long long int b,
                     long long int m = 1000000007) {
  a %= m;
  long long int res = 1;
  while (b) {
    if (b & 1) res = (res * a) % m;
    a = (a * a) % m;
    b >>= 1;
  }
  return res;
}
long long int bpow(long long int a, long long int b) {
  long long int res = 1;
  while (b) {
    if (b & 1) res = res * a;
    a = a * a;
    b >>= 1;
  }
  return res;
}
long long int modinv(long long int a, long long int m = 1000000007) {
  return modpow(a, m - 2, m);
}
void graph(vector<vector<int> > &adj, int m) {
  int x, y;
  for (int i = 0; i < m; i++) {
    cin >> x >> y;
    adj[--x].push_back(--y);
    adj[y].push_back(x);
  }
}
void solve() {
  int n;
  cin >> n;
  if (n <= 3) {
    cout << -1 << '\n';
    return;
  }
  int temp = n;
  while (temp >= 8) {
    cout << temp - 1 << " " << temp - 3 << " " << temp << " " << temp - 2
         << " ";
    temp -= 4;
  }
  if (temp == 4) {
    cout << "3 1 4 2" << '\n';
  } else if (temp == 5) {
    cout << "5 2 4 1 3" << '\n';
  } else if (temp == 6) {
    cout << "5 3 6 2 4 1" << '\n';
  } else {
    cout << "5 1 3 6 2 4 7" << '\n';
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t = 1;
  cin >> t;
  while (t--) solve();
}
