Ternary XOR

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 ci=(ai+bi)%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≤104) — the number of test cases. Then t test cases follow. The first line of the test case contains one integer n (1≤n≤5⋅104) — 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⋅104 (∑n≤5⋅104).

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

题意：是输入一个x，这个x长度为n，里面的数只有0，1，2组成，同时也给出了a,b.(a,b也是有长度为n，里面的数有0，1，2组成)其中（ai+bi)%3=xi;然后去找min(max(a,b));
思路：用贪心去做，由于题中给出了x；然后遍历x的每一个字符，就三种情况，1）为0，a,b同时加0；
2）为1，a加1，b加0；此时就出现了max由于找的是min(max),则maxa后面都要为0，其余都加在b上面。
3）为2.a,b同时加1；
上代码

package DIC3C;

import java.util.Scanner;

public class I {
     
	public static void main(String[] args) {
     
		Scanner sc = new Scanner(System.in);
		int t=sc.nextInt();
		while(t-->0) {
     
			int n=sc.nextInt();
			String str=sc.next();
			StringBuffer str1=new StringBuffer("1");//字符串的增加，主要用StringBuffer
			StringBuffer str2=new StringBuffer("1");
			for(int i=1;i<str.length();i++) {
     
				if(str.charAt(i)=='1') {
     //主要char，不要写成1，要写成’1‘
					str1.append(str.charAt(i));
					str2.append(0);
					for(int j=i+1;j<n;j++) {
     
						str1.append(0);
						str2.append(str.charAt(j));
					}
					break;
				}else if(str.charAt(i)=='2') {
     
					str1.append('1');
					str2.append('1');
				}else if(str.charAt(i)=='0') {
     
					str1.append('0');
					str2.append('0');
				}
			}
			System.out.println(str1);
			System.out.println(str2);
		}
	}
}