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.PrintWriter;
import java.util.Scanner;

public class C1328 {
    public static void main(String[] args){
        Scanner in= new Scanner(System.in);
        PrintWriter out =new PrintWriter(System.out);
        int t= in.nextInt();
        String number;
        for (int z = 0; z< t ; z++) {
            int n= in.nextInt();
            StringBuilder num1, num2;
            num1= new StringBuilder();
            num2= new StringBuilder();
            number=in.next();
            char temp;
            num1.append('1');
            num2.append('1');
            int biggest=0;
            for (int i = 1; i < n; i++) {
                temp=number.charAt(i);
                if(temp=='2'){
                    if (biggest==0) {
                        num1.append('1');
                        num2.append('1');
                    }
                    if (biggest==1){
                        num1.append('0');
                        num2.append('2');
                    }
                }
                else if (temp=='1'){
                    if(biggest==0){
                        num1.append('1');
                        num2.append('0');
                        biggest=1;
                    }
                    else if (biggest==1){
                        num1.append('0');
                        num2.append('1');
                    }
                }
                else {
                    num1.append('0');
                    num2.append('0');
                }
            }
            out.println(num1);
            out.println(num2);
            out.flush();
        }
    }
}
