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
from functools import reduce
import operator
import itertools
from bisect import bisect_left
import functools
from itertools import cycle


# n, k = map(int, input().split(' '))
# a = map(int, input().split(' '))
# d = list(map(lambda v: (v[0], sum(1 for _ in v[1])), itertools.groupby(sorted(a))))
# print(d)

def unpack(func):
    return lambda iterator: func(*iterator)

def vzip(iterator,value):
    return zip(iterator,cycle([value]))
    
def tryOr(exp,value):
    try:
        return exp()
    except Exception:
        return value



def solve():
    l = int(input())
    n = list(map(int,input()))
    p = l

    p = tryOr(lambda: n.index(1),l)
    fp = list(map(unpack(operator.floordiv),vzip(n,2)))
    
    print(*(fp[:p] + ([1] if p != l else []) + [0]*(l - p - 1)),sep='')
    print(*(fp[:p] + ([0] if p != l else []) + n[p+1:]),sep='')
    # print(p)

t = int(input())
for i in range(t):
    solve()