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.util.ArrayList;
import java.util.Arrays;
import java.util.BitSet;
import java.util.StringTokenizer;

public class F {

    static ArrayList<Integer> ans = new ArrayList<>();
    private static void solve() throws Exception {
        int n = fs.nextInt();
        char[] a = fs.next().toCharArray();
        char[] b = fs.next().toCharArray();

        Range r = new Range(0, n-1);
        r.a=a;
        ans.clear();
        n --;
        while (n >= 0) {
            if (b[n] == r.getLast()) {
                n --;
                r.removeLast();
            } else if (b[n] == other(r.getFirst())) {
                ans.add(n+1);
                r.op(n);
                r.removeLast();
                n--;
            } else {
                ans.add(1);
                ans.add(n+1);
                r.flipFirst();
                r.op(n);
                r.removeLast();
                n--;
            }
        }

        out.print(ans.size());
        out.print(' ');
        for (int x : ans) {
            out.print(x);
            out.print(' ');
        }
        out.println();
    }

    private static class Range {
        int s, e;
        boolean reversed;
        boolean flipped;
        public char[] a;

        Range(int a, int b) { s = a; e = b;}

        public char getFirst() {
            if (reversed) {
                return flipped ? other(a[e]) : a[e];
            } else {
                return flipped ? other(a[s]) : a[s];
            }
        }

        public char getLast() {
            if (reversed) {
                return flipped ? other(a[s]) : a[s] ;
            } else {
                return flipped ? other(a[e]) : a[e];
            }
        }

        public void flipFirst() {
            if (reversed) {
                a[e] = other(a[e]);
            } else {
                a[s] = other(a[s]);
            }
        }

        public void removeLast() {
            if (reversed) {
                s ++;
            } else {
                e --;
            }
        }

        public void op(int n) {
            flipped = !flipped;
            reversed = !reversed;
        }
    }

    private static void op(int n, char[] a) {
        ans.add(n);
        for (int i = 0 ; i < n ; i ++ ) { a[i] = other(a[i]); }
        int s=0, e=n-1;
        while (s<e) {
            char temp = a[s];
            a[s] = a[e];
            a[e] = temp;
            s++;
            e--;
        }
    }
    private static char other(char c) {
        return c == '0' ? '1' : '0';
    }

    private static FastScanner fs = new FastScanner();
    private static PrintWriter out = new PrintWriter(System.out);
    public static void main(String[] args) throws  Exception{
        int T = fs.nextInt();
        for (int t = 0; t < T; t ++ ) {
            solve();
        }
        out.close();
    }

    static void debug(Object... O) {
        System.out.print("DEBUG ");
        System.out.println(Arrays.deepToString(O));
    }

    static class FastScanner {
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st=new StringTokenizer("");
        String next() {
            while (!st.hasMoreTokens())
                try {
                    st=new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            return st.nextToken();
        }

        int nextInt() {
            return Integer.parseInt(next());
        }
        int[] readArray(int n) {
            int[] a=new int[n];
            for (int i=0; i<n; i++) a[i]=nextInt();
            return a;
        }
        long nextLong() {
            return Long.parseLong(next());
        }

        double nextDouble() {
            return Double.parseDouble(next());
        }

        String nextString() {
            return next();
        }
    }
}
