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
# https://codeforces.com/problemset/problem/1328/C
# inp = [2,2,0,2,2,2,0,2,1]

count = int(input())

while count > 0:

    inp_count = int(input())
    inp = int(input())
    lst_1 = ['1']
    lst_2 = ['1']
    str_inp = str(inp)


    for e in str_inp[1:]:
        if e == '2':
            lst_1.append('1')
            lst_2.append('1')
        elif e == '0':
            lst_1.append('0')
            lst_2.append('0')
        elif e == '1':
            lst_1.append('1')
            lst_2.append('0')
            break
    
    for e in str_inp[len(lst_1):]:
        lst_1.append('0')
        lst_2.append(e)

    # if '1' not in str_inp:
    #     num_1 = int(inp/2)
    #     num_2 = num_1
    #     print(num_1)
    #     print(num_2)
    # else:
    #     f = 0
    #     idx = str_inp.index('1')
    #     lst_1a = str_inp[0:idx]
    #     lst_2a = str_inp[idx+1:]
    #     if lst_1a:
    #         f = int(int(lst_1a)/2)
    #         lst_1.append(str(f))
    #         lst_1.append('1')
    #         lst_2.append(str(f))
    #         lst_2.append('0')
    #         d = inp_count - len(''.join(lst_1))
    #         if d > 0:
    #             lst_1.append(d*'0')
    #         lst_2.append(lst_2a)

    print(int(''.join(lst_1)))
    print(int(''.join(lst_2)))

    count -= 1
