human
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
Using python can you solve the prior task?
assistant
#Here is Python program 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.

# Problem : C. Ternary XOR
# Contest : Codeforces Round #629 (Div. 3)
# URL : https://codeforces.com/contest/1328/problem/C
# Memory Limit : 256 MB
# Time Limit : 1000 ms
# Powered by CP Editor (https://github.com/cpeditor/cpeditor)

"""
// Author : snape_here - Susanta Mukherjee
     
 """
 
from __future__ import division, print_function
 
import os,sys
from io import BytesIO, IOBase
 
if sys.version_info[0] < 3:
    from __builtin__ import xrange as range
    from future_builtins import ascii, filter, hex, map, oct, zip
 
 
def ii(): return int(input())
def si(): return input()
def mi(): return map(int,input().split())
def li(): return list(mi())
 
 
def read():
    sys.stdin = open('input.txt', 'r')  
    sys.stdout = open('output.txt', 'w') 
 
def gcd(x, y):
    while y:
        x, y = y, x % y
    return x

mod=1000000007

import math

def main():

    for _ in range(ii()):
    	n=ii()
    	s=si()
    	c="1"
    	d="1"
    	f=1
    	for i in range(1,n):
    		if s[i]=='1':
    			if f:
	    			c+='0'
	    			d+='1'
	    			f-=1
	    		else:
	    			c+='1'
	    			d+='0'
    		elif s[i]=='2':
    			if f:
    				c+='1'
    				d+='1'
    			else:
	    			c+='2'
	    			d+='0'
    		else:
    			c+='0'
    			d+='0'
    	print(c)
    	print(d)
    
# region fastio
 
BUFSIZE = 8192
 
 
class FastIO(IOBase):
    newlines = 0
 
    def __init__(self, file):
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None
 
    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()
 
    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size,...