This is the easy version of the problem. The difference between the versions is the constraint on n and the required number of operations. You can make hacks only if all versions of the problem are solved.

There are two binary strings a and b of length n (a binary string is a string consisting of symbols 0 and 1). In an operation, you select a prefix of a, and simultaneously invert the bits in the prefix (0 changes to 1 and 1 changes to 0) and reverse the order of the bits in the prefix.

For example, if a=001011 and you select the prefix of length 3, it becomes 011011. Then if you select the entire string, it becomes 001001.

Your task is to transform the string a into b in at most 3n operations. It can be proved that it is always possible.

Input

The first line contains a single integer t (1≤ t≤ 1000) — the number of test cases. Next 3t lines contain descriptions of test cases.

The first line of each test case contains a single integer n (1≤ n≤ 1000) — the length of the binary strings.

The next two lines contain two binary strings a and b of length n.

It is guaranteed that the sum of n across all test cases does not exceed 1000.

Output

For each test case, output an integer k (0≤ k≤ 3n), followed by k integers p_1,…,p_k (1≤ p_i≤ n). Here k is the number of operations you use and p_i is the length of the prefix you flip in the i-th operation.

Example

Input


5
2
01
10
5
01011
11100
2
01
01
10
0110011011
1000110100
1
0
1


Output


3 1 2 1
6 5 2 5 3 1 2
0
9 4 1 2 10 4 1 2 1 5
1 1

Note

In the first test case, we have 01→ 11→ 00→ 10.

In the second test case, we have 01011→ 00101→ 11101→ 01000→ 10100→ 00100→ 11100.

In the third test case, the strings are already the same. Another solution is to flip the prefix of length 2, which will leave a unchanged.
import java.io.*;
import java.math.BigInteger;
import java.util.*;

public class Main {
    BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    PrintWriter pw = new PrintWriter(System.out);
    static int MOD = 1000000007;

    public static void main(String[] args) throws IOException {
        Main m = new Main();
        m.solve();
        m.close();
    }

    void close() throws IOException {
        pw.flush();
        pw.close();
        br.close();
    }

    int ri() throws IOException {
        return Integer.parseInt(br.readLine());
    }

    long rl() throws IOException {
        return Long.parseLong(br.readLine());
    }

    int[] ril(int n) throws IOException {
        int[] nums = new int[n];
        int c = 0;
        for (int i = 0; i < n; i++) {
            int sign = 1;
            c = br.read();
            int x = 0;
            if (c == '-') {
                sign = -1;
                c = br.read();
            }
            while (c >= '0' && c <= '9') {
                x = x * 10 + c - '0';
                c = br.read();
            }
            nums[i] = x * sign;
        }
        while (c != '\n' && c != -1) c = br.read();
        return nums;
    }

    long[] rll(int n) throws IOException {
        long[] nums = new long[n];
        int c = 0;
        for (int i = 0; i < n; i++) {
            int sign = 1;
            c = br.read();
            long x = 0;
            if (c == '-') {
                sign = -1;
                c = br.read();
            }
            while (c >= '0' && c <= '9') {
                x = x * 10 + c - '0';
                c = br.read();
            }
            nums[i] = x * sign;
        }
        while (c != '\n' && c != -1) c = br.read();
        return nums;
    }

    char[] rs() throws IOException {
        return br.readLine().toCharArray();
    }

    void solve() throws IOException {
        int t = ri();
        for (int ti = 0; ti < t; ti++) {
            int n = ri();
            char[] a = rs();
            char[] b = rs();
            List<Integer> ans = new ArrayList<>();
            for (int i = n-1; i >= 0; i--) {
                if (a[i] != b[i]) {
                    if (a[0] == b[i]) {
                        ans.add(0);
                        a[0] = a[0] == '0' ? '1' : '0';
                    }
                    reverseAndFlip(a, 0, i);
                    ans.add(i);
                }
            }
            pw.print(ans.size());
            for (int x : ans) pw.print(" " + (x+1));
            pw.println();
        }
    }

    void reverseAndFlip(char[] a, int i, int j) {
        while (i <= j) {
            char tmp = a[i];
            a[i] = a[j];
            a[j] = tmp;
            a[i] = a[i] == '0' ? '1' : '0';
            if (i != j) a[j] = a[j] == '0' ? '1' : '0';
            i++;
            j--;
        }
    }
}