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 POSITIVE_INFINITY = 9223372036854775807;
const long long NEGATIVE_INFINITY = -9223372036854775807;
const long long MOD = 1000000007;
const long double PI = acos(-1.0);
template <typename T>
inline T bigMod(T A, T B, T M = MOD) {
  A %= M;
  T ret = 1;
  while (B) {
    if (B & 1) ret = (ret * A) % M;
    A = (A * A) % M;
    B = B >> 1;
  }
  return ret;
}
template <typename T>
inline T invMod(T A, T M = MOD) {
  return bigMod(A, M - 2, M);
}
const int N = 200005;
void solve() {
  int n;
  cin >> n;
  if (n <= 3) {
    cout << -1 << "\n";
    return;
  }
  vector<int> ar;
  if (n % 2 == 0) {
    for (int i = 2; i <= n; i += 2) {
      ar.push_back(i);
    }
    for (int i = n - 1; i >= 1; i -= 2) {
      ar.push_back(i);
    }
    int m = n / 2;
    swap(ar[m], ar[m + 1]);
  } else {
    for (int i = 1; i <= n; i += 2) {
      ar.push_back(i);
    }
    for (int i = n - 1; i > 1; i -= 2) {
      ar.push_back(i);
    }
    int m = n / 2;
    swap(ar[m + 1], ar[m + 2]);
  }
  for (int v : ar) {
    cout << v << " ";
  }
  cout << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  int tc;
  cin >> tc;
  while (tc--) {
    solve();
  }
  return 0;
}
