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;
void swap(long long int &a, long long int &b) {
  long long int tmp = a;
  a = b;
  b = tmp;
}
long long int atoistring(string s) {
  stringstream lol(s);
  long long int x;
  lol >> x;
  return x;
}
void pv(vector<long long int> v) {
  for (long long int i = 0; i < v.size(); i++) {
    cout << v[i] << " ";
  }
  cout << "\n";
}
void pa(long long int *v, long long int size) {
  for (long long int i = 0; i < size; i++) {
    cout << v[i] << " ";
  }
  cout << "\n";
}
long long int power(long long int x, long long int y, long long int p) {
  long long int res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) {
      res = (res * x) % p;
    }
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
void checkmod(long long int &x, long long int mod) {
  if (x >= mod) {
    x %= mod;
  }
}
long long int sumofgp(long long int a, long long int n, long long int m) {
  if (n < 0) {
    return 0;
  }
  if (n == 0) {
    return 1;
  }
  if (n == 1) {
    return (1 + a) % m;
  }
  long long int ans;
  if (n % 2) {
    ans = (1 + a) * sumofgp((a * a) % m, (n - 1) / 2, m);
  } else {
    ans = 1 + ((a * (1 + a)) % m) * sumofgp((a * a) % m, n / 2 - 1, m);
  }
  return ans % m;
}
long long int power_of_2_not_less(long long int n) {
  if (!n) {
    return 1;
  }
  n--;
  n |= n >> 1;
  n |= n >> 2;
  n |= n >> 4;
  n |= n >> 8;
  n |= n >> 16;
  n |= n >> 32;
  n++;
  return n;
}
bool done[1003];
int main(void) {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int t;
  cin >> t;
  while (t--) {
    long long int n;
    cin >> n;
    for (long long int i = 0; i < n + 1; i++) done[i] = 0;
    if (n < 4) {
      cout << "-1\n";
      continue;
    }
    if (n == 4) {
      cout << "3 1 4 2\n";
      continue;
    }
    long long int i;
    for (i = n - n % 2; i > 0; i -= 2) {
      cout << i << " ";
      done[i] = 1;
    }
    if (i == 0) {
      cout << "5 ";
      done[5] = 1;
    } else {
      cout << "4 ";
      done[4] = 1;
    }
    for (i = 1; i <= n; i++) {
      if (done[i]) continue;
      done[i] = 1;
      cout << i << " ";
    }
    cout << "\n";
  }
  return 0;
}
