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
from collections import defaultdict,deque
import sys
input=sys.stdin.readline
def bfs(i):
    queue=deque([(i,1)])
    while queue:
        #print(queue)
        vertex,colour=queue.popleft()
        res.append(vertex)
        if colour==1:
            up,down=a[vertex],b[vertex]
        else:
            up,down=b[vertex],a[vertex]
        for child in graph[vertex]:
            if not vis[child]:
                vis[child]=True
                if a[child]==up or b[child]==down:
                    tmp.append(child)
                    queue.append((child,0))
                else:
                    queue.append((child,1))
    
t=int(input())
for ii in range(t):
    n=int(input())
    a=[int(i) for i in input().split() if i!='\n']
    b=[int(i) for i in input().split() if i!='\n']
    data=[[] for _ in range(n)]
    for i in range(n):
        data[a[i]-1].append(i)
        data[b[i]-1].append(i)
    graph=defaultdict(list)
    for i in range(n):
        if len(data[i])!=2:
            print(-1)
            break
    else:
        for i in range(n):
            u,v=data[i][0],data[i][1]
            graph[u].append(v)
            graph[v].append(u)
        #print(graph,data)
        vis,ans=[False]*(n),[]
        for i in range(n):
            if not vis[i]:
                tmp,res=[],[]
                vis[i]=True
                bfs(i)
                if len(res)-len(tmp)>=len(tmp):
                    ans+=tmp
                else:
                    tmp=set(tmp)
                    for i in res:
                        if i not in tmp:
                            ans.append(i)
        print(len(ans))
        ans=[i+1 for i in ans]
        print(*ans)
        
            
        
    
                   
    
    
