Codeforces Round #629 (Div. 3) C. Ternary XOR（贪心）

A number is ternary if it contains only digits 00 , 11 and 22 . For example, the following numbers are ternary: 10221022 , 1111 , 2121 , 20022002 .

You are given a long ternary number xx . The first (leftmost) digit of xx is guaranteed to be 22 , the other digits of xx can be 00 , 11 or 22 .

Let's define the ternary XOR operation ⊙⊙ of two ternary numbers aa and bb (both of length nn ) as a number c=a⊙bc=a⊙b of length nn , where ci=(ai+bi)%3ci=(ai+bi)%3 (where %% is modulo operation). In other words, add the corresponding digits and take the remainders of the sums when divided by 33 . For example, 10222⊙11021=2121010222⊙11021=21210 .

Your task is to find such ternary numbers aa and bb both of length nn and both without leading zeros that a⊙b=xa⊙b=x and max(a,b)max(a,b) is the minimum possible.

You have to answer tt independent test cases.

Input

The first line of the input contains one integer tt (1≤t≤1041≤t≤104 ) — the number of test cases. Then tt test cases follow. The first line of the test case contains one integer nn (1≤n≤5⋅1041≤n≤5⋅104 ) — the length of xx . The second line of the test case contains ternary number xx consisting of nn digits 0,10,1 or 22 . It is guaranteed that the first digit of xx is 22 . It is guaranteed that the sum of nn over all test cases does not exceed 5⋅1045⋅104 (∑n≤5⋅104∑n≤5⋅104 ).

Output

For each test case, print the answer — two ternary integers aa and bb both of length nn and both without leading zeros such that a⊙b=xa⊙b=x and max(a,b)max(a,b) is the minimum possible. If there are several answers, you can print any.

Example

Input

Copy

4
5
22222
5
21211
1
2
9
220222021

Output

Copy

11111
11111
11000
10211
1
1
110111011
110111010
题干啰里啰唆一堆==
观察一下样例甚至都能找到贪心策略：因为要尽可能保证大的那个数较小，所以尽可能使两个数接近。设置一个标记big，从原数的最高位往低位遍历，当big没有被赋值时，如果原数这一位是2：新数的这一位都赋为1；如果原数这一位是0：新数的这一位都赋为0；如果原数这一位是1：随便给一个数的这一位赋1，并把big设置为这个数，另一个数赋值为0；只要big标记设置了，以后原数的每一位都给非big标记的数，有big标记的数赋值为0.
emmm还是看代码吧。

#include <bits/stdc++.h>
using namespace std;
char s[50005],s1[50005],s2[50005];
int n;
int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        cin>>n;
        scanf("%s",s);
        s1[0]='1';s2[0]='1';
        int i;
        int big=-1;
        for(i=1;i<=n-1;i++)
        {
            if(big==1)//已经能分出谁大谁小了 
            {
                s2[i]=s[i];
                s1[i]='0';
            }
            else
            {
                if(s[i]=='0')
                {
                    s1[i]=s2[i]='0';
                }
                else if(s[i]=='1')
                {
                    s1[i]='1';
                    s2[i]='0';
                    big=1;
                }
                else if(s[i]=='2')
                {
                    s1[i]=s2[i]='1';
                }
            }
        }
        for(i=0;i<n;i++)
        {
            cout<<s1[i];
        }
        cout<<endl;
        for(i=0;i<n;i++)
        {
            cout<<s2[i];
        }
        cout<<endl;
    }
    return 0;
}