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 A {
	static FastReader sc=null;
	
	public static void main(String[] args) {
		sc=new FastReader();
		PrintWriter out=new PrintWriter(System.out);
		int t=sc.nextInt();
		while(t-->0) {
			int n=sc.nextInt();
			char B[]=sc.next().toCharArray();
			char A[]=new char[n];
			 A[0] = '1';
			 
			    for (int i = 1; i < n; i++) {
			        A[i] = '1';
			 
			        if (A[i] + B[i] == A[i - 1] + B[i - 1])
			            A[i] = '0';
			    }
			    for(char e:A)out.print(e);
			    out.println();
			    
		}
		out.close();
		
		
		
		
		
		
	}
	
	
	
	
	
	
	
	static int[] reverse(int a[]) {
		ArrayList<Integer> al=new ArrayList<>();
		for(int i:a)al.add(i);
		Collections.sort(al,Collections.reverseOrder());
		for(int i=0;i<a.length;i++)a[i]=al.get(i);
		return a;
	}
	static int gcd(int a,int b) {
		if(b==0)return a;
		else return gcd(b,a%b);
	}
	static long gcd(long a,long b) {
		if(b==0)return a;
		else return gcd(b,a%b);
	}
	
	static void ruffleSort(int a[]) {
		ArrayList<Integer> al=new ArrayList<>();
		for(int i:a)al.add(i);
		Collections.sort(al);
		for(int i=0;i<a.length;i++)a[i]=al.get(i);
	}
	
	
	static void print(long a[]) {
		for(long e:a) {
			System.out.print(e+" ");
		}
		System.out.println();
	}
	static void print(char a[]) {
		for(char e:a) {
			System.out.print(e+" ");
		}
		System.out.println();
	}
	
	
	static void print(int a[]) {
		for(int e:a) {
			System.out.print(e+" ");
		}
		System.out.println();
	}
	static void print(double a[]) {
		for(double e:a) {
			System.out.print(e+" ");
		}
		System.out.println();
	}
	
	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; 
        } 
        int[] readArray(int n) {
    		int a[]=new int [n];
    		for(int i=0;i<n;i++) {
    			a[i]=sc.nextInt();
    		}
    		return a;
    	}
    } 
}






