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;
#pragma GCC target("avx2")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int long long t, ic;
  cin >> t;
  for (int ic = 1; ic <= t; ic++) {
    int long long num;
    cin >> num;
    if (num == 2 || num == 3) {
      cout << -1 << '\n';
      continue;
    }
    if (num == 4) {
      cout << "3 1 4 2" << '\n';
      continue;
    }
    vector<int long long> arr(num);
    if (num % 2 == 0) {
      int long long i = num - 1;
      int long long j = 0;
      while (1) {
        int long long val = num - (2 * j);
        if (val == 0) {
          break;
        } else {
          arr[i] = val;
        }
        i--;
        j++;
      }
      int long long c = num - 1;
      j = 0;
      while (1) {
        if (c == 5) {
          c = c - 2;
        }
        arr[j] = c;
        if (c == -1) {
          arr[j] = 5;
          break;
        }
        c = c - 2;
        j++;
      }
    } else {
      int long long i = num - 1;
      int long long j = num - 1;
      while (1) {
        if (i == 0) {
          break;
        }
        arr[j] = i;
        i = i - 2;
        j--;
      }
      int long long c = num;
      j = 0;
      while (1) {
        if (c == 5) {
          c = c - 2;
        }
        arr[j] = c;
        if (c == -1) {
          arr[j] = 5;
          break;
        }
        c = c - 2;
        j++;
      }
    }
    for (int i = 0; i < num; i++) {
      cout << arr[i] << " ";
    }
    cout << '\n';
  }
  return 0;
}
