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>
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma GCC optimize("unroll-loops")
#pragma GCC optimize("Ofast")
#pragma GCC optimize("O3")
#pragma GCC target("avx")
const long long MOD = 1000000007;
const long double PI = 3.1415926535898;
const long long LIM = 100005;
using namespace std;
long long fpow(long long x, long long y) {
  long long temp;
  if (y == 0) return 1;
  temp = fpow(x, y / 2);
  if (y % 2 == 0)
    return temp * temp;
  else
    return x * temp * temp;
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
void sieve(long long n) {
  bool prime[5 * LIM];
  memset(prime, true, sizeof(prime));
  for (long long p = 2; p * p <= n; p++)
    if (prime[p] == true)
      for (long long i = p * p; i <= n; i += p) prime[i] = false;
  prime[1] = 0;
}
inline long long read() {
  long long p = 1, ans = 0;
  char ch = getchar();
  while (ch < 48 || ch > 57) {
    if (ch == '-') p = -1;
    ch = getchar();
  }
  while (ch >= 48 && ch <= 57) {
    ans = (ans << 3) + (ans << 1) + ch - 48;
    ch = getchar();
  }
  return ans * p;
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t;
  cin >> t;
  for (long long tt = 1; tt <= t; ++tt) {
    long long n;
    cin >> n;
    if (n < 4) {
      cout << -1 << endl;
      continue;
    }
    long long ele = 0;
    if (n % 2 == 0)
      ele = n - 1;
    else
      ele = n;
    while (ele >= 1) {
      cout << ele << " ";
      ele -= 2;
    }
    cout << 4 << " " << 2 << " ";
    ele = 6;
    while (ele <= n) {
      cout << ele << " ";
      ele += 2;
    }
    cout << endl;
  }
  return 0;
}
