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.io.*;
import java.util.*;
public class cf
{
   // static class pair{
   //  long x;
   //  Long y;
   //  public pair(long x,long y){
   //      this.x=x;
   //      this.y=y;
   //  }
   // }
   static boolean prime[];
  static  void prime(int n) 
    { 
      
        prime= new boolean[n+1]; 
        for(int i=0;i<n;i++) 
            prime[i] = true; 
          
        for(int p = 2; p*p <=n; p++) 
        { 
            // If prime[p] is not changed, then it is a prime 
            if(prime[p] == true) 
            { 
                // Update all multiples of p 
                for(int i = p*p; i <= n; i += p) 
                    prime[i] = false; 
            } 
        } 
        
    } 

      
    public static void main(String args[])
    {
      BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
      Scanner sc=new Scanner(System.in);
     prime(100000);
       int t=sc.nextInt();
       while(t-->0){
        int n=sc.nextInt();
        String s=sc.next();
        char ans[]=new char[100001];
        ans[0]='1';
        int k=s.charAt(0)+1;
        char p=(char)k;
        for(int i=0;i<n;i++){
           if(s.charAt(i)+'1'!=p){
            ans[i]='1';
           }   
           else{
            ans[i]='0';
           }
           k=s.charAt(i)+ans[i];
           p=(char)k;
        }
        for(int i=0;i<n;i++){
          System.out.print(ans[i]);   
        }
        System.out.println();

       }

   }
}