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.Scanner;

public class Puzzle {
	public static void main(String[] args) {
			int t;
			Scanner sc = new Scanner(System.in);
			t = sc.nextInt();
			while (t > 0) {
				int n = sc.nextInt();
				String b = sc.next();
				StringBuilder sb=new StringBuilder();
				//System.out.println(S);
				int prevsum=0;
				if(b.charAt(0)=='1') {
					prevsum=2;
				}
				else if(b.charAt(0)=='0') {
					prevsum=1;
				}
				int newdig=1;
				sb.append(newdig);
				int ns=0;
				//System.out.println("sss"+prevsum);
				int c=1;
				while(c<n) {
					//System.out.println(n);
					if(prevsum==1 && b.charAt(c)=='0') {
						ns=0;
						newdig=0;
					}
					if(prevsum==1 && b.charAt(c)=='1') {
						ns=2;
						newdig=1;
					}
					if(prevsum==2 && b.charAt(c)=='1') {
						ns=1;
						newdig=0;
					}
					if(prevsum==2 && b.charAt(c)=='0') {
						ns=1;
						newdig=1;
					}
					if(prevsum==0 && b.charAt(c)=='1') {
						ns=2;
						newdig=1;
					}
					if(prevsum==0 && b.charAt(c)=='0') {
						ns=1;
						newdig=1;
					}
					prevsum=ns;
					sb.append(newdig);
					//System.out.println(sb.toString());
					c++;
				}
				System.out.println(sb.toString());
				t--;
			}
	}
}
