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.io.*;
import java.math.BigDecimal;
import java.math.BigInteger;
import java.util.*;

public class Main {
    static PrintWriter out;
    static int  n, m, maxn = 1;
    static String ss;
    static LinkedList<Integer> linkedList = new LinkedList<>();
    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        out = new PrintWriter(new BufferedOutputStream(System.out));
        int t = Integer.parseInt(br.readLine().trim());
        while(t-->0){
            String S[] = br.readLine().split(" ");
            int n = Integer.parseInt(S[0]);
            List<Integer> list = new ArrayList<>();
            if(n < 4){
                out.println(-1);
            }else if(n == 4){
                out.println(3+" "+1+" "+4+" "+2);
            }else {
                int m = n;
                if(m % 2 == 0){
                    while (m > 6){
                        list.add(m);
                        m -= 2;
                    }
                    list.add(6);
                    list.add(3);
                    list.add(1);
                    list.add(4);
                    list.add(2);
                    list.add(5);
                    m+=1;
                    while(m < n){
                        list.add(m);
                        m += 2;
                    }
                }else {
                    while (m > 5){
                        list.add(m);
                        m -= 2;
                    }
                    list.add(5);
                    list.add(3);
                    list.add(1);
                    list.add(4);
                    list.add(2);
                    m+=1;
                    while(m < n){
                        list.add(m);
                        m += 2;
                    }
                }
            }
            for(int i = 0; i < list.size();i++){
                out.print(list.get(i)+" ");
            }
            out.println();
        }

        out.close();
    }



    static class MyUtils {
        //《---------------------------工具类---------------------------》\

        //从大到小排序
        static Comparator<Integer> cmp = new Comparator<Integer>() {
            public int compare(Integer i1, Integer i2) {
                return i2-i1;
            }
        };

        //[begin, end) 进行二分查找，返回 大于或等于 tar的第一个元素位置。如果所有元素都小于tar，则返回 end.
        public static int lowerBound(int []nums,int l,int r,int target){
            while(l<r){
                int m = (l+r)/2;
                if(nums[m]<=target) l = m+1;
                else    r = m;
            }
            return l;
        }
        //全排列
        public static void perm(int[] array,int start,int end) {
            if(start==end) {
                System.out.println(Arrays.toString(array));
            } else {
                for (int i = start; i <= end; i++) {
                    //1，2，3的全排列这块相当于将其中一个提了出来，下次递归从start+1开始
                    swap(array,start,i);
                    perm(array,start+1,end);
                    //这块是复原数组，为了保证下次另外的同级递归使用数组不会出错
                    //这块可以通过树来理解，每次回退一步操作，交换回去
                    swap(array,start,i);
                }
            }
        }
        //交换两数
        public static void swap(int[] array,int i,int j) {
            int temp = array[i];
            array[i] = array[j];
            array[j] = temp;
        }

        private static int gcd(int a, int b) {
            int t ;
            if(a < b){
                t=a;
                a=b;
                b=t;
            }
            return a%b==0?b:(gcd(b,a%b));
        }

        // [begin, end) 进行二分查找，返回 大于 tar的第一个元素位置。如果所有元素都小于tar，则返回 end.
        public static int upperBound(int []nums ,int l,int r, int target){
            while(l<r){
                int m = (l+r)/2;
                if(nums[m]>=target) r= m;
                else    l = m +1;
            }
            return l;
        }
    }

    static class Pairs {
        int first;
        int second;
        public Pairs(int first, int second) {
            this.first = first;
            this.second = second;
        }
    }
}



