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.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.util.Scanner;
import java.util.StringTokenizer;

public class PuzzleFrmFut {
	   public 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;
	        }
	    }

	public static void main(String[] args) throws IOException {
		// TODO Auto-generated method stub
		FastReader sc = new FastReader();
        int t=sc.nextInt();
        
		while(t-- > 0) {
	        int n = sc.nextInt();
	        String b = sc.nextLine();
	        char a[] = new char[n];
//	        String a = "";
			String d = "";
			int start = 0;
			int prev = -1;
			for(int i = 0 ; i < n ; i++) {
				//we dont need d
				//can use a pointer last_sum for keeping track of last sum of a + b at ith index
				if(start == 0) {
//					a += "1";
					a[i] = '1';
					if(b.charAt(i) == '1') {
//						d+="2";
						prev = 2;
					}else {
//						d+="1";
						prev = 1;
					}
					start = 1;
				}
				else {
					if(b.charAt(i) == '1') {
						if(prev != 2) {
//							a+="1";
							a[i] = '1';
							prev = 2;
						}else {
//							a+="0";
							a[i] = '0';
//							d+="1";
							prev = 1;
						}
					}
					else {
						if(prev != 1) {
//							a+="1";
							a[i] = '1';
//							d+="1";
							prev = 1;
						}
						else {
//							a+="0";
							a[i] = '0';
//							d+="0";
							prev = 0;
						}
					}
				}
			}
			System.out.println(a);
		}
	}

}
