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 MOD = 1e9 + 7;
const int N = 1e3 + 100;
const int INF = 0x3f3f3f3f;
const long long LINF = 0x3f3f3f3f3f3f3f3f;
inline long long dpow(long long a, long long b) {
  long long r = 1, t = a;
  while (b) {
    if (b & 1) r = (r * t) % MOD;
    b >>= 1;
    t = (t * t) % MOD;
  }
  return r;
}
inline long long fpow(long long a, long long b) {
  long long r = 1, t = a;
  while (b) {
    if (b & 1) r = (r * t);
    b >>= 1;
    t = (t * t);
  }
  return r;
}
vector<int> v;
int main() {
  int T;
  cin >> T;
  while (T--) {
    int n;
    scanf("%d", &n);
    v.clear();
    if (n < 4)
      printf("-1\n");
    else {
      int j = 1;
      while (j <= n) v.push_back(j), j += 2;
      j = ((n - 1) / 2 + 1) * 2;
      if (j > n) j -= 2;
      while (j >= 2) v.push_back(j), j -= 2;
      if (n & 1)
        swap(v[(n - 1) / 2 + 1], v[(n - 1) / 2 + 1 + 1]);
      else
        swap(v[n / 2 - 2], v[n / 2 - 1]);
      for (auto it : v) cout << it << " ";
      cout << endl;
    }
  }
  return 0;
}
