A number is ternary if it contains only digits 0, 1 and 2. For example, the following numbers are ternary: 1022, 11, 21, 2002.

You are given a long ternary number x. The first (leftmost) digit of x is guaranteed to be 2, the other digits of x can be 0, 1 or 2.

Let's define the ternary XOR operation ⊙ of two ternary numbers a and b (both of length n) as a number c = a ⊙ b of length n, where c_i = (a_i + b_i) \% 3 (where \% is modulo operation). In other words, add the corresponding digits and take the remainders of the sums when divided by 3. For example, 10222 ⊙ 11021 = 21210.

Your task is to find such ternary numbers a and b both of length n and both without leading zeros that a ⊙ b = x and max(a, b) is the minimum possible.

You have to answer t independent test cases.

Input

The first line of the input contains one integer t (1 ≤ t ≤ 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 ≤ 5 ⋅ 10^4) — the length of x. The second line of the test case contains ternary number x consisting of n digits 0, 1 or 2. It is guaranteed that the first digit of x is 2. It is guaranteed that the sum of n over all test cases does not exceed 5 ⋅ 10^4 (∑ n ≤ 5 ⋅ 10^4).

Output

For each test case, print the answer — two ternary integers a and b both of length n and both without leading zeros such that a ⊙ b = x and max(a, b) is the minimum possible. If there are several answers, you can print any.

Example

Input


4
5
22222
5
21211
1
2
9
220222021


Output


11111
11111
11000
10211
1
1
110111011
110111010
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.util.Random;
import java.util.StringTokenizer;

public class ProblemB {

public static Scanner sc = new Scanner(System.in);

  
	// This code is contributed by PrinciRaj1992

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










public static void solve() throws IOException {
	
	
	int count =sc.nextInt();
	for(int i=0;i<count;i++) {
		boolean isOne=false;
		StringBuilder first=new StringBuilder("");
		StringBuilder second=new StringBuilder("");
		StringBuilder sb =  new StringBuilder("");
		sc.nextLine();
		sb.append(sc.nextLine());
		String s=sb.toString();
		for(int j=0;j<s.length();j++) {
			if(s.charAt(j)=='2' && !isOne) {
				first.append("1");
				second.append("1");
			}
			else if(s.charAt(j)=='0' && !isOne) {
				first.append("0");
				second.append("0");
			}
			else if(s.charAt(j)=='1' && !isOne) {
				first.append("1");
				second.append("0");
				isOne=true;
			}
			else {
				first.append("0");
				second.append(s.charAt(j)+"");
				
			}
		}
		System.out.println(first);
		System.out.println(second);
		
		
		
		
		
		
		
	}

}
}
 class Scanner 
{
	StringTokenizer st;
	BufferedReader br;

	public Scanner(InputStream s){	br = new BufferedReader(new InputStreamReader(s));}

	public String next() throws IOException 
	{
		while (st == null || !st.hasMoreTokens()) 
			st = new StringTokenizer(br.readLine());
		return st.nextToken();
	}

	public int nextInt() throws IOException {return Integer.parseInt(next());}
	
	public long nextLong() throws IOException {return Long.parseLong(next());}

	public String nextLine() throws IOException {return br.readLine();}
	
	public double nextDouble() throws IOException
	{
		String x = next();
		StringBuilder sb = new StringBuilder("0");
		double res = 0, f = 1;
		boolean dec = false, neg = false;
		int start = 0;
		if(x.charAt(0) == '-')
		{
			neg = true;
			start++;
		}
		for(int i = start; i < x.length(); i++)
			if(x.charAt(i) == '.')
			{
				res = Long.parseLong(sb.toString());
				sb = new StringBuilder("0");
				dec = true;
			}
			else
			{
				sb.append(x.charAt(i));
				if(dec)
					f *= 10;
			}
		res += Long.parseLong(sb.toString()) / f;
		return res * (neg?-1:1);
	}
	
	public boolean ready() throws IOException {return br.ready();}


}




























