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;
int dx[] = {0, 0, 1, -1, -1, -1, 1, 1};
int dy[] = {1, -1, 0, 0, -1, 1, 1, -1};
char ch[] = {'N', 'S', 'E', 'W'};
template <class T>
inline T biton(T n, T pos) {
  return n | ((T)1 << pos);
}
template <class T>
inline T bitoff(T n, T pos) {
  return n & ~((T)1 << pos);
}
template <class T>
inline T ison(T n, T pos) {
  return (bool)(n & ((T)1 << pos));
}
template <class T>
inline T gcd(T a, T b) {
  while (b) {
    a = a % b;
    swap(a, b);
  }
  return a;
}
template <class T>
inline T bigmod(T p, T e, T m) {
  T ret = 1;
  for (; e > 0; e >>= 1) {
    if (e & 1) ret = (ret * p) % m;
    p = (p * p) % m;
  }
  return (T)ret;
};
long long n;
bool possible(long long dif) {
  vector<long long> res;
  set<long long> st;
  for (int j = 1; j <= 2; j++) {
  }
  for (int i = 1; i <= n; i = i + dif) {
    st.insert(i);
    res.push_back(i);
  }
  if (st.size() == n) {
    return true;
  }
}
void solve() {
  cin >> n;
  if (n <= 3) {
    cout << -1 << "\n";
    ;
    return;
  }
  vector<long long> res;
  for (int i = (n / 2 * 2); i >= 6; i = i - 2) {
    res.push_back(i);
  }
  res.push_back(2);
  res.push_back(4);
  res.push_back(1);
  res.push_back(3);
  for (int i = 5; i <= n; i = i + 2) {
    res.push_back(i);
  }
  for (int i = 1; i < n; i++) {
    int d = abs(res[i] - res[i - 1]);
    if (d >= 2 && d <= 4) {
      continue;
    } else {
      cout << -1 << "\n";
      ;
      return;
    }
  }
  for (int i = 0; i < res.size(); i++) {
    cout << res[i] << " ";
  }
  cout << "\n";
  ;
}
int main() {
  ios_base::sync_with_stdio(false);
  cout.setf(ios::fixed);
  cout.precision(10);
  cout.tie(nullptr);
  cin.tie(nullptr);
  int t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}
