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;
long long pr[1000010], fact[1000010];
void prime() {
  pr[0] = 1;
  pr[1] = 1;
  for (long long i = 2; i < sqrt(1000007); i++) {
    for (long long j = 2 * i; j <= 1000007; j += i) {
      pr[j] = 1;
    }
  }
}
void facto() {
  long long i = 1;
  fact[0] = 1;
  while (i <= 1000007) {
    fact[i] = (fact[i - 1] * i) % 1000000007;
    i++;
  }
}
long long fastpowmd(long long n, long long r) {
  if (r == 0) return 1;
  if (r == 1) return n;
  long long hr = fastpowmd(n, r / 2);
  if (r % 2 == 0)
    return (hr * hr) % 1000000007;
  else
    return (((hr * hr) % 1000000007) * n) % 1000000007;
}
long long fastpow(long long n, long long r) {
  if (r == 0) return 1;
  long long hr = fastpow(n, r / 2);
  if (r % 2 == 0)
    return (hr * hr);
  else
    return (hr * hr * n);
}
int countDivisors(long long n) {
  long long cnt = 0;
  for (long long i = 1; i <= sqrt(n); i++) {
    if (n % i == 0) {
      if (n / i == i)
        cnt++;
      else
        cnt = cnt + 2;
    }
  }
  return cnt;
}
int inv(long long n) { return fastpowmd(n, 1000000007 - 2); }
long long gcd(long long a, long long b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
long long multmod(long long n, long long m) {
  return ((n % 1000000007) * (m % 1000000007)) % 1000000007;
}
long long addmod(long long n, long long m) {
  return ((n % 1000000007) + (m % 1000000007)) % 1000000007;
}
long long si(string second) {
  stringstream shrey(second);
  long long x = 0;
  shrey >> x;
  return x;
}
string is(long long n) {
  string str = to_string(n);
  return str;
}
long long bi(string second) {
  reverse(second.begin(), second.end());
  long long val = 0;
  for (long long i = 0; i < second.length(); i++) {
    val += (fastpow(2, i) * (second[i] - '0'));
  }
  return val;
}
long long ncr(long long n, long long r) {
  long long ans = 1;
  for (long long i = 1; i <= r; i += 1) {
    ans = multmod(ans, (n - r + i));
    long long deno = inv(i);
    ans = multmod(ans, deno);
  }
  return ans;
}
long long nd(long long N, long long R) { return ncr(N - 1, R - 1); }
long long a[1000005];
vector<long long> g[100005];
long long mat[2001][2001], visited[2001][2001], weight[2001][2001];
vector<pair<long long, pair<long long, long long> > > parent[21][21];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    if (n == 1)
      cout << 1 << "\n";
    else if (n == 2 || n == 3)
      cout << -1 << "\n";
    else if (n == 4)
      cout << "3 1 4 2\n";
    else if (n == 5)
      cout << "3 1 4 2 5\n";
    else {
      if (n & 1) {
        for (long long i = 1; i <= n; i += 2) cout << i << " ";
        cout << n - 3 << " " << n - 1 << " ";
        for (long long i = n - 5; i > 0; i -= 2) cout << i << " ";
        cout << "\n";
      } else {
        for (long long i = 2; i <= n; i += 2) cout << i << " ";
        cout << n - 3 << " " << n - 1 << " ";
        for (long long i = n - 5; i > 0; i -= 2) cout << i << " ";
        cout << "\n";
      }
    }
  }
  return 0;
}
