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 CodeForcesTest {
	public static void main(String[] args) throws IOException {
        FastReader sc=new FastReader();
        BufferedWriter out = new BufferedWriter(new OutputStreamWriter(new FileOutputStream(java.io.FileDescriptor.out), "ASCII"), 512);
		int t=sc.nextInt();
        while(t--!=0){
			int n=sc.nextInt();
			String s = sc.next();
			char[] c = new char[n];
			char[] ans = new char[n];
			if(s.charAt(0)=='0'){
				c[0]='1';
			}
			else{
				c[0]='2';
			}
			ans[0]='1';
			
			for(int i=1;i<n;i++){
				if(s.charAt(i)=='1' && c[i-1]=='2'){
					c[i]='1';
					ans[i]='0';
				}
				else if(s.charAt(i)=='1' && c[i-1]=='1'){
					c[i]='2';
					ans[i]='1';
				}
				else if(s.charAt(i)=='0' && c[i-1]=='1'){
					c[i]='0';
					ans[i]='0';
				}
				else if(s.charAt(i)=='0' && c[i-1]=='2'){
					c[i]='1';
					ans[i]='1';
				}
				else if(s.charAt(i)=='1' && c[i-1]=='0'){
					c[i]='2';
					ans[i]='1';
				}
				else if(s.charAt(i)=='0' && c[i-1]=='0'){
					c[i]='1';
					ans[i]='1';
				}
			}
			System.out.println(ans);
        }
        out.close();
	}
    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; 
		} 
	} 
	static int Mex(int[] a){
        Arrays.sort(a);
        if(a[0]!=1){
            return 1;
        }
        for(int i=0;i<a.length-1;i++){
            if(a[i]!=a[i+1] && a[i+1]>a[i]+1){
                return a[i]+1;
            }
        }
        return a[a.length-1]+1;
    }
	static int gcd(Integer a,Integer b)
    {
        if(b==0)
            return a;
        else
            return gcd(b,a%b);
    }
	static int[] readArray(int n){
		FastReader sc = new FastReader();
		int[] a = new int[n];
		for(int i=0;i<n;i++){
			a[i]=sc.nextInt();
		}
		return a;
	}
}
