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.
// Akash Yadav(aky91)

import java.util.*;
import java.lang.*;
import java.io.*;
import java.math.*;

public class Solution{

	public static Scanner scn = new Scanner(System.in);
	public static boolean multipleTC = true;

	public static void main (String[] args) throws java.lang.Exception{
		int T = multipleTC?scn.nextInt():1;
		while(T-- > 0) solve();
	}

	public static void solve(){
	    
	    int n = scn.nextInt();
	    
	    String a = scn.next();
	    String b = scn.next();
	    
	    if(a.equals(b)){
	        pln(0);
	        return;
	    }
	    
	    int flip = 0;
	    ArrayList<Integer> ops = new ArrayList<>();
	    boolean allZero = (a.charAt(0)=='0');
	    
	    //pln(flip + " " + allZero);
	    
	    //make all bits of a same
	    for(int i = 1; i < n; i++){
	        if(a.charAt(i) == '0'){
	            if(allZero){
	                continue;
	            } else {
	                flip++;
	                ops.add(i-1);
	                allZero = true;
	            }
	        } else {
	            if(!allZero){
	                continue;
	            } else {
	                flip++;
	                ops.add(i-1);
	                allZero = false;
	            }
	        }
	    }
	    
	    //pln(flip + " " + allZero);
	    
	    //make a same as b
	    for(int i = n-1; i >= 0; i--){
	        if(b.charAt(i) == '0'){
	            if(allZero){
	                continue;
	            } else {
	                flip++;
	                ops.add(i);
	                allZero = true;
	            }
	        } else {
	            if(!allZero){
	                continue;
	            } else {
	                flip++;
	                ops.add(i);
	                allZero = false;
	            }
	        }
	    }
	    
	    //pln(flip + " " + allZero);
	    
	    p(flip + " ");
	    for(int v : ops) p((v+1) + " ");
	    pln();
	}

	
	public static void p(Object o){System.out.print(o + " ");}
	public static void pln(Object o){System.out.println("" + o);}
	public static void pln(){System.out.println("");}
}