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 
# cook your dish here
t=int(input())
for _ in range(t):
    n=int(input())
    if n==2 or n==3:
        print(-1)
    elif n==4:
        print(2,4,1,3)
    else:
        ans=[]
        i=1
        while i<=n:
            ans.append(i)
            i=i+2
        i=n if n%2==0 else n-1
        while i>0:
            ans.append(i)
            i=i-2
        f=0
        for i in range(n-1):
            if abs(ans[i]-ans[i+1])==1:
                if n%2==0:
                    if i+3<=n-1:
                        ans[i+1],ans[i+3]=ans[i+3],ans[i+1]
                        ans[i+2],ans[i+3]=ans[i+3],ans[i+2]
                    else:
                        f=1
                else:
                    if i+2<=n-1:
                        ans[i+1],ans[i+2]=ans[i+2],ans[i+1]
                    else:
                        f=1
                break
        if f:
            print(-1)
        else:
            print(*ans)