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 long long INF = 1e18;
const long double pi = acos(-1.0);
int gcd(int a, int b) {
  while (b) {
    a %= b;
    swap(a, b);
  }
  return a;
}
int lcm(int a, int b) { return a / gcd(a, b) * b; }
long long binpows(long long a, long long b) {
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = res * a;
    a = a * a;
    b >>= 1;
  }
  return res;
}
int binpow(int a, int b, int m) {
  a %= m;
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = res * a % m;
    a = a * a % m;
    b >>= 1;
  }
  return res;
}
const int mod = 1e9 + 7;
void solve() {
  int n;
  cin >> n;
  if (n == 1) {
    cout << 1 << "\n";
  } else if (n == 2 or n == 3) {
    cout << -1 << "\n";
  } else {
    deque<int> d;
    d.push_back(2);
    d.push_back(4);
    d.push_back(1);
    d.push_back(3);
    for (int i = 5; i <= n; i++) {
      if (i % 2 == 0) {
        d.push_front(i);
      } else {
        d.push_back(i);
      }
    }
    while (!d.empty()) {
      cout << d.front() << " ";
      d.pop_front();
    }
    cout << "\n";
  }
}
signed main() {
  cin.tie(NULL);
  cout.tie(NULL);
  ios_base::sync_with_stdio(false);
  int t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
}
