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 
import java.util.*;
import java.io.*;

public class tr0 {
	static PrintWriter out;
	static StringBuilder sb;
	static int mod = 998244353;
	static long inf = (long) 1e18 + 1;
	static int[] col;
	static int n, m;
	static TreeSet<Integer>[] ad;
	static long[] memo, cnt[];
	static HashSet<Integer> h;
	static int[] pos, ones;
	static char[][] g;
	static boolean[][] vis;
	static boolean f;

// 1 3 5 7 9 6 10 8 4 2
	public static void main(String[] args) throws Exception {
		Scanner sc = new Scanner(System.in);
		out = new PrintWriter(System.out);
		int t = sc.nextInt();
		while (t-- > 0) {
			int n = sc.nextInt();
			if (n == 2 || n == 3)
				out.println(-1);
			else {
				int[] ans = new int[n];
				if (n % 2 == 0) {
					if (n == 4) {
						ans = new int[] { 3, 1, 4, 2 };
					}
					else if (n == 6) {
						ans = new int[] { 5, 3, 6, 2, 4, 1 };
					} else {
						int id = 0;
						for (int i = 1; i < n; i += 2)
							ans[id++] = i;
						ans[id] = n - 4;
						ans[++id] = n;
						ans[++id] = n - 2;
						ans[++id] = n - 6;
						int k = n - 8;
						while (++id < n) {
							ans[id] = k;
							k -= 2;
						}
					}
				} else {
					int id = 0;
					for (int i = 1; i <= n; i += 2)
						ans[id++] = i;
					ans[id++] = n - 3;
					ans[id++] = n - 1;
					int k = n - 5;
					while (id < n) {
						ans[id++] = k;
					k-=2;
					}
				}
				for (int i = 0; i < n; i++)
					out.print(ans[i] + " ");
				out.println();
			}
		}
		out.flush();
	}

	static class Scanner {
		StringTokenizer st;
		BufferedReader br;

		public Scanner(InputStream system) {
			br = new BufferedReader(new InputStreamReader(system));
		}

		public Scanner(String file) throws Exception {
			br = new BufferedReader(new FileReader(file));
		}

		public String next() throws IOException {
			while (st == null || !st.hasMoreTokens())
				st = new StringTokenizer(br.readLine());
			return st.nextToken();
		}

		public String nextLine() throws IOException {
			return br.readLine();
		}

		public int nextInt() throws IOException {
			return Integer.parseInt(next());
		}

		public double nextDouble() throws IOException {
			return Double.parseDouble(next());
		}

		public char nextChar() throws IOException {
			return next().charAt(0);
		}

		public Long nextLong() throws IOException {
			return Long.parseLong(next());
		}

		public int[] nextArrInt(int n) throws IOException {
			int[] a = new int[n];
			for (int i = 0; i < n; i++)
				a[i] = nextInt();
			return a;
		}

		public long[] nextArrLong(int n) throws IOException {
			long[] a = new long[n];
			for (int i = 0; i < n; i++)
				a[i] = nextLong();
			return a;
		}

		public boolean ready() throws IOException {
			return br.ready();
		}

		public void waitForInput() throws InterruptedException {
			Thread.sleep(3000);
		}
	}
}