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 acmp {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int a = scanner.nextInt();
        int[] ns = new int[a];
        String[] bs = new String[a];
        String[] as = new String[a];
        String[] ds = new String[a];

        for (int i = 0; i < a; i++) {
            ns[i] = scanner.nextInt();
            bs[i] = scanner.next();

//            System.out.println("ns= " + ns[i] + " i= " + i);
//            System.out.println("ds= " + bs[i] + " i= " + i);
        }
        for (int i = 0; i < a; i++) {
            as[i] = "1";
            String prev = String.valueOf(1 + Character.getNumericValue(bs[i].charAt(0)));
//            ds[i] = String.valueOf(1 + Character.getNumericValue(bs[i].charAt(0)));
            for (int j = 1; j < ns[i]; j++) {
                if(1 + Character.getNumericValue(bs[i]
                        .charAt(j))
                        !=
                        Integer.parseInt(prev)){
                    as[i] += "1";
//                    ds[i] += String.valueOf(1 + Character.getNumericValue(bs[i].charAt(j)));
                    prev = String.valueOf(1 + Character.getNumericValue(bs[i].charAt(j)));
                }else {
                    as[i] += "0";
//                    ds[i] += bs[i].charAt(j);
                    prev = String.valueOf(bs[i].charAt(j));
                }
            }
//            System.out.println(as[i]);
//            System.out.println(ds[i]);
        }

        for (String s : as) {
            System.out.println(s);
        }
//        System.out.println();
//
//        for (String d : ds) {
//            System.out.println(d);
//        }
    }
}
