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

sys.setrecursionlimit(10 ** 5)
int1 = lambda x: int(x) - 1
p2D = lambda x: print(*x, sep="\n")
def II(): return int(sys.stdin.readline())
def MI(): return map(int, sys.stdin.readline().split())
def LI(): return list(map(int, sys.stdin.readline().split()))
def LLI(rows_number): return [LI() for _ in range(rows_number)]
def SI(): return sys.stdin.readline()[:-1]

def maketo():
    atoi=[[] for _ in range(n)]
    for i,a in enumerate(aa):
        if a<1 or a>n:return True
        atoi[a-1].append(i)
        if len(atoi[a-1])>2:return True
    for a in range(n):
        u,v=atoi[a]
        if u+n==v:continue
        to[u].append((v+n)%(n*2))
        to[v].append((u+n)%(n*2))
    return False

def dfs(u):
    flap[u%n]=(u>=n)*1
    cur[u>=n].append(u%n+1)
    stack=[u]
    while stack:
        u=stack.pop()
        for v in to[u]:
            if flap[v%n]==-1:
                flap[v % n]=(v>=n)*1
                cur[v >= n].append(v % n+1)
                stack.append(v)
            elif flap[v % n]!=(v>=n)*1:return True
    return False

for _ in range(II()):
    n=II()
    aa=LI()+LI()
    to=[[] for _ in range(n*2)]

    if maketo():
        print(-1)
        continue
    #print(to)

    ans = []
    flap=[-1]*n
    ng=False
    for u in range(n):
        #u+=n
        if flap[u%n]!=-1:continue
        cur=[[],[]]
        ng=dfs(u)
        if len(cur[0])<len(cur[1]):ans+=cur[0]
        else:ans+=cur[1]
        #print(u,flap,cur,ans)
        if ng:break

    if ng:print(-1)
    else:
        print(len(ans))
        print(*ans)
