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
#include <bits/stdc++.h>
using namespace std;
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    int n;
    scanf("%d", &n);
    vector<vector<int>> a(3, vector<int>(n + 1));
    vector<int> num(n + 1, 0);
    vector<vector<int>> pos(n + 1);
    for (int i = 1; i <= 2; i++)
      for (int j = 1; j <= n; j++) {
        scanf("%d", &a[i][j]);
        num[a[i][j]]++;
        pos[a[i][j]].push_back(j);
      }
    bool f = 1;
    for (int i = 1; i <= n; i++)
      if (num[i] != 2) {
        f = 0;
        break;
      }
    if (!f) {
      printf("-1\n");
      continue;
    }
    vector<bool> vis(n + 1, 0);
    vector<int> ans;
    for (int i = 1; i <= n; i++) {
      if (vis[i]) continue;
      if (pos[i][0] == pos[i][1]) continue;
      vector<int> tmp1, tmp2;
      for (int j = i, k = pos[i][0]; !vis[j];) {
        vis[j] = 1;
        k = k ^ pos[j][0] ^ pos[j][1];
        if (a[1][k] == j) {
          tmp1.push_back(k);
          j = a[2][k];
        } else {
          tmp2.push_back(k);
          j = a[1][k];
        }
      }
      if (tmp1.size() <= tmp2.size())
        ans.insert(ans.end(), tmp1.begin(), tmp1.end());
      else
        ans.insert(ans.end(), tmp2.begin(), tmp2.end());
    }
    printf("%d\n", (int)ans.size());
    for (auto i : ans) printf("%d ", i);
    printf("\n");
  }
  return 0;
}
