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;
const int maxn = 1e4 + 10;
vector<int> G[maxn];
int vis[maxn];
vector<int> ans, ans1;
int n, flag;
void dfs1(int u) {
  int cur = 0;
  vis[u] = 1;
  for (int i = 0; i < G[u].size(); i++) {
    if (vis[G[u][i]] == 0 && cur == 0)
      ans1.push_back(G[u][i]), dfs1(G[u][i]), cur = 1;
  }
}
void dfs(int u, int tot) {
  if (tot == n) {
    for (int i = 0; i < ans.size(); i++) cout << ans[i] << " ";
    cout << endl;
    flag = 1;
    return;
  }
  for (int i = 0; i < G[u].size(); i++) {
    if (flag) break;
    if (vis[G[u][i]] == 0 && flag == 0) {
      ans.push_back(G[u][i]), vis[G[u][i]] = 1, dfs(G[u][i], tot + 1),
                              ans.pop_back(), vis[G[u][i]] = 0;
    }
  }
}
int main() {
  std::ios::sync_with_stdio(false);
  std::cin.tie(0);
  int T;
  cin >> T;
  while (T--) {
    cin >> n;
    ans.clear();
    ans1.clear();
    memset(vis, 0, sizeof(vis));
    if (n <= 3) {
      cout << "-1" << endl;
      continue;
    }
    for (int i = 1; i <= n; i++) G[i].clear();
    for (int i = 1; i <= n; i++) {
      for (int j = i + 2; j <= i + 4 && j <= n; j++) {
        G[i].push_back(j), G[j].push_back(i);
      }
    }
    vis[2] = 1;
    flag = 0;
    ans.push_back(2);
    dfs(2, 1);
    if (flag == 0)
      cout << "-1" << endl;
    else {
    }
  }
  return 0;
}
