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.InputStreamReader;
import java.util.Scanner;


public class c3
{


    public void solve(String x)
    {
       
        StringBuilder sa=new StringBuilder("");        
        StringBuilder sb=new StringBuilder("");

        

        int big=0;
        int index=0;

       
        while(index!=x.length())
        {
            if(x.charAt(index)=='2')
            {
                if(big==0)
                {
                sa.append("1");
                sb.append("1");
                }
                else if(big==1)
                {
                    sa.append("0");
                    sb.append("2");
                }
                else
                {
                    sa.append("2");
                    sb.append("0");
                }
                index++;
            }

            else if(x.charAt(index)=='1')
            {
                if(big==0)
                {
                sa.append("1");
                sb.append("0");
                big=1;
                }
                else if(big==1)
                {
                    sa.append("0");
                    sb.append("1");
                }
                else
                {
                    sa.append("1");
                    sb.append("0");
                }
                index++;
            }
            else
            {
                sa.append("0");
                sb.append("0");
                index++;
            }
        }

        System.out.println(sa.toString());
        System.out.println(sb.toString());
    }

    public static void main(String args[]) throws Exception
    {
        c3 c=new c3();
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
        int t=Integer.parseInt(br.readLine());
        String x="";
       
        while(t-->0)
        {
            String temp=br.readLine();
            x=br.readLine();
            c.solve(x);
        }

    }

    

}