In the 2022 year, Mike found two binary integers a and b of length n (both of them are written only by digits 0 and 1) that can have leading zeroes. In order not to forget them, he wanted to construct integer d in the following way: 

  * he creates an integer c as a result of bitwise summing of a and b without transferring carry, so c may have one or more 2-s. For example, the result of bitwise summing of 0110 and 1101 is 1211 or the sum of 011000 and 011000 is 022000; 
  * after that Mike replaces equal consecutive digits in c by one digit, thus getting d. In the cases above after this operation, 1211 becomes 121 and 022000 becomes 020 (so, d won't have equal consecutive digits). 



Unfortunately, Mike lost integer a before he could calculate d himself. Now, to cheer him up, you want to find any binary integer a of length n such that d will be maximum possible as integer.

Maximum possible as integer means that 102 > 21, 012 < 101, 021 = 21 and so on.

Input

The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.

The first line of each test case contains the integer n (1 ≤ n ≤ 10^5) — the length of a and b.

The second line of each test case contains binary integer b of length n. The integer b consists only of digits 0 and 1.

It is guaranteed that the total sum of n over all t test cases doesn't exceed 10^5.

Output

For each test case output one binary integer a of length n. Note, that a or b may have leading zeroes but must have the same length n.

Example

Input


5
1
0
3
011
3
110
6
111000
6
001011


Output


1
110
100
101101
101110

Note

In the first test case, b = 0 and choosing a = 1 gives d = 1 as a result.

In the second test case, b = 011 so: 

  * if you choose a = 000, c will be equal to 011, so d = 01; 
  * if you choose a = 111, c will be equal to 122, so d = 12; 
  * if you choose a = 010, you'll get d = 021. 
  * If you select a = 110, you'll get d = 121. 

We can show that answer a = 110 is optimal and d = 121 is maximum possible.

In the third test case, b = 110. If you choose a = 100, you'll get d = 210 and it's the maximum possible d.

In the fourth test case, b = 111000. If you choose a = 101101, you'll get d = 212101 and it's maximum possible d.

In the fifth test case, b = 001011. If you choose a = 101110, you'll get d = 102121 and it's maximum possible d.
import java.util.*;
import java.io.*;
 
public class codeforces
{
    public static void main(String[] args) 
    {
        FastReader s=new FastReader();
        int t=s.nextInt();

        while(t-->0)
        {
            int n=s.nextInt();
            String a=s.next();

            String output="";
            int prev=0;

            if(a.charAt(0)=='0')
            {
                output+='1';
                prev=1;
            }
            else
            {
                output+='1';
                prev=2;
            }

            for(int i=1;i<n;i++)
            {
                if(a.charAt(i)=='0')
                {
                    if(prev==0)
                    {
                        output+='1';
                        prev=1;
                    }
                    else if(prev==1)
                    {
                        output+='0';
                        prev=0;
                    }
                    else{
                        output+='1';
                        prev=1;
                    }
                }
                else
                {
                    if(prev==0)
                    {
                        output+='1';
                        prev=2;
                    }
                    else if(prev==1)
                    {
                        output+='1';
                        prev=2;
                    }
                    else 
                    {
                        output+='0';
                        prev=1;
                    }
                }
            }
            System.out.println(output);
        }
    }

    public static boolean isPrime(int num)
    {
        for(int i=2;i<=Math.sqrt(num);i++)
            if(num%i==0)
                return false;
        return true;
    }
 
    //FastReader//////////////////////////////////////////////////////////////////////////////
    private static class FastReader
    {
        BufferedReader br;
        StringTokenizer st;
 
        public FastReader()
        {
            br = new BufferedReader(new
                     InputStreamReader(System.in));
        }
 
        String next()
        {
            while (st == null || !st.hasMoreElements())
            {
                try
                {
                    st = new StringTokenizer(br.readLine());
                }
                catch (IOException  e)
                {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }
 
        int nextInt()
        {
            return Integer.parseInt(next());
        }
 
        long nextLong()
        {
            return Long.parseLong(next());
        }
 
        double nextDouble()
        {
            return Double.parseDouble(next());
        }
 
        String nextLine()
        {
            String str = "";
            try
            {
                str = br.readLine();
            }
            catch (IOException e)
            {
                e.printStackTrace();
            }
            return str;
        }
    }
    
}