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
input=sys.stdin.readline
def read_list():
	return map(int,input().split())
def print_list(l):
	print(' '.join(map(str,l)))
# import heapq as hq
# import bisect as bs
# from collections import deque as dq
# from collections import defaultdict as dc 
# import math

# f = open('test.py')
# def input():
# 	return f.readline().replace('\n','')

def solve(n,A):
	find = [-1]*(n+1)
	same = [0]*(2*n)
	for i in range(2*n):
		num = A[i]
		if find[num] == -1:
			find[num] = i 
		elif find[num]!=-2:
			same[i] = find[num]
			same[find[num]] = i 
			find[num] = -2
		else:
			return None
	ans = []
	see = [False]*(2*n)
	for root in range(2*n):
		if see[root]:
			continue
		node = root 
		c0 = []
		c1 = []
		while not see[node]:
			see[node] = True
			see[same[node]] = True
			if node&1:
				c1.append(node>>1)
			else:
				c0.append(node>>1)
			node = same[node^1]
		if len(c0)<len(c1):
			ans+=c0 
		else:
			ans+=c1
	return ans		

data = list(map(int,sys.stdin.read().split()))
out = []
N = data[0]
tt = 1
for _ in range(N):
	n = data[tt]
	tt += 1
	a = data[tt:tt+n]
	tt += n
	b = data[tt:tt+n]
	tt += n
	nums = [a[i>>1] if i&1==0 else b[i>>1] for i in range(2*n)]
	ans = solve(n,nums)
	if ans is None:
		# print(-1)
		out.append('-1')
	else:
		out.append(str(len(ans)))
		out.append(' '.join(map(str,[x+1 for x in ans])))
print('\n'.join(out))