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 java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.StringTokenizer;

public class Task7 {
    boolean[] visited;
    int[][] ar;
    int[] colourOf;
    int[][] occurences;

    public static void main(String[] args) throws IOException {
        new Task7().solve();
    }

    private void solve() throws IOException {
        BufferedReader f = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter out = new PrintWriter(System.out);

        int t = Integer.parseInt(f.readLine());
        for (int t1 = 0; t1 < t; t1++) {
            int n = Integer.parseInt(f.readLine());
            ar = new int[2][n];
            occurences = new int[n + 1][2];

            int[] count = new int[n + 1];
            for (int i = 0; i < 2; i++) {
                StringTokenizer tokenizer = new StringTokenizer(f.readLine());
                for (int j = 0; j < n; j++) {
                    ar[i][j] = Integer.parseInt(tokenizer.nextToken());
                    count[ar[i][j]]++;
                    if (occurences[ar[i][j]][0] == 0) {
                        occurences[ar[i][j]][0] = j;
                    } else {
                        occurences[ar[i][j]][1] = j;
                    }
                }
            }

            boolean impossible = false;
            for (int i = 1; i <= n; i++) {
                if (count[i] != 2) {
                    impossible = true;
                    break;
                }
            }
            if (impossible) {
                out.println(-1);
                continue;
            }

            visited = new boolean[n];
            colourOf = new int[n];

            ArrayList<Integer> solution = new ArrayList<Integer>();
            for (int i = 0; i < n; i++) {
                if (!visited[i] && ar[0][i] != ar[1][i]) {
                    ArrayList<Integer> colour0 = new ArrayList<Integer>();
                    ArrayList<Integer> colour1 = new ArrayList<Integer>();
                    boolean possible = tryToColour(0, i, colour0, colour1);
                    if (!possible) {
                        impossible = true;
                        break;
                    }

                    if (colour0.size() < colour1.size()) {
                        solution.addAll(colour0);
                    } else {
                        solution.addAll(colour1);
                    }
                }

            }
            if (impossible) {
                out.println(-1);
                continue;
            }

            out.println(solution.size());
            if (!solution.isEmpty()) {
                out.print(solution.get(0) + 1);
                for (int i = 1; i < solution.size(); i++) {
                    out.print(" ");
                    out.print(solution.get(i) + 1);
                }
            }
            out.println();
        }


        out.close();
    }

    private boolean tryToColour(int colour, int col, ArrayList<Integer> colour0, ArrayList<Integer> colour1) {
        //System.out.println(col + " " + colour);
        if (visited[col]) {
            if (colour == colourOf[col]) return true;
            else return false;
        }

        visited[col] = true;
        boolean possible = true;
        colourOf[col] = colour;
        if (colour == 0) colour0.add(col);
        else  colour1.add(col);

        for (int i = 0; i < 2; i++) {
            int otherCol = occurences[ar[i][col]][0];
            if (otherCol == col) otherCol = occurences[ar[i][col]][1];
            if (ar[i][otherCol] == ar[i][col]) possible = possible && tryToColour(otherColour(colour), otherCol, colour0, colour1);
            else possible = possible && tryToColour(colour, otherCol, colour0, colour1);
        }

        return possible;
    }

    private int otherColour(int colour) {
        if (colour == 1) return 0;
        return 1;
    }
}
