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 optimize("Ofast")
int ah = 911382323;
int bh = 972663749;
template <typename F, typename S>
ostream &operator<<(ostream &os, const pair<F, S> &p) {
  return os << "(" << p.first << ", " << p.second << ")";
}
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &v) {
  os << "{";
  typename vector<T>::const_iterator it;
  for (it = v.begin(); it != v.end(); it++) {
    if (it != v.begin()) os << ", ";
    os << *it;
  }
  return os << "}";
}
template <typename T>
ostream &operator<<(ostream &os, const set<T> &v) {
  os << "[";
  typename set<T>::const_iterator it;
  for (it = v.begin(); it != v.end(); it++) {
    if (it != v.begin()) os << ", ";
    os << *it;
  }
  return os << "]";
}
template <typename F, typename S>
ostream &operator<<(ostream &os, const map<F, S> &v) {
  os << "[";
  typename map<F, S>::const_iterator it;
  for (it = v.begin(); it != v.end(); it++) {
    if (it != v.begin()) os << ", ";
    os << it->first << " = " << it->second;
  }
  return os << "]";
}
long long dx[8] = {1, -1, 0, 0, 1, 1, -1, -1};
long long dy[8] = {0, 0, 1, -1, 1, -1, 1, -1};
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long(t);
  cin >> (t);
  while ((t)--) {
    long long n;
    cin >> n;
    if (n == 2 || n == 3)
      cout << (-1) << " ";
    else if (n % 2 == 0) {
      vector<long long> ans;
      long long x = 1;
      while (x < n - 3) {
        ans.emplace_back(x);
        x += 2;
      }
      x += 2;
      ans.emplace_back(x);
      x -= 2;
      ans.emplace_back(x);
      x = n;
      while (x > 0) {
        ans.emplace_back(x);
        x -= 2;
      }
      for (auto x : ans) cout << (x) << " ";
    } else {
      vector<long long> ans;
      long long x = 1;
      while (x <= n) {
        ans.emplace_back(x);
        x += 2;
      }
      x -= 2;
      x -= 3;
      ans.emplace_back(x);
      x += 2;
      ans.emplace_back(x);
      x -= 4;
      while (x > 0) {
        ans.emplace_back(x);
        x -= 2;
      }
      for (auto x : ans) cout << (x) << " ";
    }
    cout << "\n";
  }
}
