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.*;
import java.util.*;

import java.math.*;
 

public class Main {
	
	static class FastReader 
    { 
        BufferedReader br; 
        StringTokenizer st; 
  
        public FastReader() 
        { 
            br = new BufferedReader(new
                     InputStreamReader(System.in)); 
        } 
  
        String next() 
        { 
            while (st == null || !st.hasMoreElements()) 
            { 
                try
                { 
                    st = new StringTokenizer(br.readLine()); 
                } 
                catch (IOException  e) 
                { 
                    e.printStackTrace(); 
                } 
            } 
            return st.nextToken(); 
        } 
  
        int nextInt() 
        { 
            return Integer.parseInt(next()); 
        } 
  
        long nextLong() 
        { 
            return Long.parseLong(next()); 
        } 
  
        double nextDouble() 
        { 
            return Double.parseDouble(next()); 
        } 
  
        String nextLine() 
        { 
            String str = ""; 
            try
            { 
                str = br.readLine(); 
            } 
            catch (IOException e) 
            { 
                e.printStackTrace(); 
            } 
            return str; 
        } 
    } 
	
	static int t;
	static String n, x;
	
	public static void main(String args[]) throws Exception {
		FastReader cin = new FastReader();

		t = cin.nextInt();
		for (int i = 0; i < t; i++) {
			n = cin.next();
			x = cin.next();
			compute(n, x);
		}
	}
	
	public static void compute(String n, String s) {
		String aAns = "";
		String bAns = "";
		
		int index = 0;
		for (int i = 0; i < s.length(); i++) {
			int temp = Character.getNumericValue(s.charAt(i));
			int a, b;
			if (temp % 2 == 0) {
				a = temp / 2;
				b = temp / 2;
				aAns += Integer.toString(a);
				bAns += Integer.toString(b);
			} else {
				a = temp;
				b = 0;
				aAns += Integer.toString(a);
				bAns += Integer.toString(b);
				System.out.print(aAns);
				for (int j = i + 1; j < s.length(); j++) {
					System.out.print(0);
				}
				System.out.println();
				System.out.print(bAns);
				System.out.println(s.substring(i + 1));
				return;
			}
		}
		System.out.println(aAns);
		System.out.println(bAns);
	}
}