You are given a table a of size 2 × n (i.e. two rows and n columns) consisting of integers from 1 to n.

In one move, you can choose some column j (1 ≤ j ≤ n) and swap values a_{1, j} and a_{2, j} in it. Each column can be chosen no more than once.

Your task is to find the minimum number of moves required to obtain permutations of size n in both first and second rows of the table or determine if it is impossible to do that.

You have to answer t independent test cases.

Recall that the permutation of size n is such an array of size n that contains each integer from 1 to n exactly once (the order of elements doesn't matter).

Input

The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.

The first line of the test case contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of columns in the table. The second line of the test case contains n integers a_{1, 1}, a_{1, 2}, ..., a_{1, n} (1 ≤ a_{1, i} ≤ n), where a_{1, i} is the i-th element of the first row of the table. The third line of the test case contains n integers a_{2, 1}, a_{2, 2}, ..., a_{2, n} (1 ≤ a_{2, i} ≤ n), where a_{2, i} is the i-th element of the second row of the table.

It is guaranteed that the sum of n does not exceed 2 ⋅ 10^5 (∑ n ≤ 2 ⋅ 10^5).

Output

For each test case print the answer: -1 if it is impossible to obtain permutation of size n in both first and the second rows of the table, or one integer k in the first line, where k is the minimum number of moves required to obtain permutations in both rows, and k distinct integers pos_1, pos_2, ..., pos_k in the second line (1 ≤ pos_i ≤ n) in any order — indices of columns in which you need to swap values to obtain permutations in both rows. If there are several answers, you can print any.

Example

Input


6
4
1 2 3 4
2 3 1 4
5
5 3 5 1 4
1 2 3 2 4
3
1 2 1
3 3 2
4
1 2 2 1
3 4 3 4
4
4 3 1 4
3 2 2 1
3
1 1 2
3 2 2


Output


0

2
2 3 
1
1 
2
3 4 
2
3 4 
-1
import sys
range = xrange;input = raw_input
inp = [int(x) for x in sys.stdin.read().split()]; ii = 0;out = []
def solve(n, A):
    same = [-1]*(2 * n);pos = [-1] * n
    for i in range(2 * n):
        a = A[i]
        if pos[a] == -1:pos[a] = i
        elif pos[a] == -2:return None
        else:p = pos[a];pos[a] = -2;same[p] = i;same[i] = p    
    ans = [];found = [0] * (2 * n)
    for root in range(2 * n):
        if found[root]:continue        
        count0 = [];count1 = [];node = root
        while not found[node]:
            found[node] = 1;found[same[node]] = 1            
            if node & 1:count0.append(node >> 1)
            else:count1.append(node >> 1)            
            node = same[node ^ 1]
        if len(count0) < len(count1):ans += count0
        else:ans += count1
    return ans
t = inp[ii]; ii += 1
for _ in range(t):
    n = inp[ii]; ii += 1
    A = [x - 1 for x in inp[ii: ii + n]]; ii += n
    B = [x - 1 for x in inp[ii: ii + n]]; ii += n    
    A = [A[i >> 1] if i & 1 == 0 else B[i >> 1] for i in range(2 * n)]
    ans = solve(n, A)
    if ans is None:out.append('-1')
    else:out.append(str(len(ans)));out.append(' '.join(str(x + 1) for x in ans))
print '\n'.join(out)