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.
/*
⠀⠀⠀⠀⣠⣶⡾⠏⠉⠙⠳⢦⡀⠀⠀⠀⢠⠞⠉⠙⠲⡀⠀
⠀⠀⠀⣴⠿⠏⠀⠀⠀⠀⠀⠀⢳⡀⠀⡏⠀⠀Y⠀⠀⢷
⠀⠀⢠⣟⣋⡀⢀⣀⣀⡀⠀⣀⡀⣧⠀⢸⠀⠀A⠀⠀ ⡇
⠀⠀⢸⣯⡭⠁⠸⣛⣟⠆⡴⣻⡲⣿⠀⣸⠀⠀S⠀  ⡇
⠀⠀⣟⣿⡭⠀⠀⠀⠀⠀⢱⠀⠀⣿⠀⢹⠀⠀H⠀⠀ ⡇
⠀⠀⠙⢿⣯⠄⠀⠀⠀⢀⡀⠀⠀⡿⠀⠀⡇⠀⠀⠀⠀⡼
⠀⠀⠀⠀⠹⣶⠆⠀⠀⠀⠀⠀⡴⠃⠀⠀⠘⠤⣄⣠⠞⠀
⠀⠀⠀⠀⠀⢸⣷⡦⢤⡤⢤⣞⣁⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀
⠀⠀⢀⣤⣴⣿⣏⠁⠀⠀⠸⣏⢯⣷⣖⣦⡀⠀⠀⠀⠀⠀⠀
⢀⣾⣽⣿⣿⣿⣿⠛⢲⣶⣾⢉⡷⣿⣿⠵⣿⠀⠀⠀⠀⠀⠀
⣼⣿⠍⠉⣿⡭⠉⠙⢺⣇⣼⡏⠀⠀⠀⣄⢸⠀⠀⠀⠀⠀⠀
⣿⣿⣧⣀⣿………⣀⣰⣏⣘⣆⣀⠀⠀
 */
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter; // System.out is a PrintWriter
// import java.util.Arrays;
// import java.util.ArrayList;
// import java.util.Collections;    //  for ArrayList sorting mainly
// import java.util.HashMap;
// import java.util.HashSet;
// import java.util.Random;
import java.util.StringTokenizer;

public class A {
    public static void main(String[] args) throws IOException {
        FastScanner scn = new FastScanner();
        PrintWriter out = new PrintWriter(System.out);
        for (int tc = scn.nextInt(); tc > 0; tc--) {
            int N = scn.nextInt();
            char[] str = scn.next().toCharArray();
            char[] ans = new char[N];
            ans[0] = '1';
            for (int i = 1; i < N; i++) {
                int val = ans[i - 1] + str[i - 1] - '0' - '0';
                int curr = str[i] - '0';
                ans[i] = (curr + 1 == val) ? '0' : '1';
            }
            for (char c : ans) {
                out.print(c);
            } out.println();
        }
        out.close();
    }

    static int gcd(int a, int b) {
        if (b == 0) {
            return a;
        }
        return gcd(b, a % b);
    }

    static class FastScanner {
        BufferedReader br;
        StringTokenizer st;

        FastScanner() {
            this.br = new BufferedReader(new InputStreamReader(System.in));
            this.st = new StringTokenizer("");
        }

        String next() {
            while (!st.hasMoreTokens()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException err) {
                    err.printStackTrace();
                }
            }
            return st.nextToken();
        }

        String nextLine() {
            if (st.hasMoreTokens()) {
                return st.nextToken("").trim();
            }
            try {
                return br.readLine().trim();
            } catch (IOException err) {
                err.printStackTrace();
            }
            return "";
        }

        int nextInt() {
            return Integer.parseInt(next());
        }

    }
}
