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;
template <typename T>
inline T S(T a) {
  return a * a;
}
template <class T>
inline string tostring(T a) {
  ostringstream os("");
  os << a;
  return os.str();
}
template <typename T>
inline long long tolong(T a) {
  long long res;
  istringstream os(a);
  os >> res;
  return res;
}
template <typename T>
inline T gcd(T a, T b) {
  if (b == 0)
    return a;
  else
    return gcd(b, a % b);
}
template <typename T>
inline T bigmod(T a, T b, T m) {
  if (b == 0)
    return 1;
  else if (b % 2 == 0)
    return S(bigmod(a, b / 2, m)) % m;
  else
    return (a % m * bigmod(a, b - 1, m)) % m;
}
const int inf = (int)1e9 + 5;
const long long linf = (long long)1e16 + 5;
const long long modd = (long long)1e9 + 7;
const int mod = 10000007;
void pr(int f) {
  if (f == 1)
    cout << "YES\n";
  else
    cout << "NO\n";
}
bool isprime(long long n) {
  if (n < 2) return 0;
  if (n == 2)
    return 1;
  else {
    for (int i = 2; i * i <= n; i++)
      if (n % i == 0) return 0;
  }
  return 1;
}
map<int, int> mp;
set<int> st;
multiset<string>::iterator it;
deque<char> q;
int vis[10000000];
pair<int, int> a[2020];
int main() {
  ios_base::sync_with_stdio(0);
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    if (n <= 3)
      cout << -1 << endl;
    else {
      for (int i = n - (n % 2 == 0); i > 0; i -= 2) cout << i << " ";
      cout << "4 2 ";
      for (int i = 6; i <= n; i += 2) cout << i << " ";
      cout << endl;
    }
  }
}
