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.InputStreamReader;
import java.util.*;


public class Main {
    private static FastScanner sc = new FastScanner();
    private static long mod = 1000000007;

    public static void main(String[] args) {
        int t = 0;
        if (sc.hasNext())
            t = sc.nextInt();
        for (int i = 1; i <= t; i++) {
            solve();
        }
    }

    public static void solve() {
        // start solving
        int n = sc.nextInt();
        char[] a = sc.next().toCharArray();
        char[] b = new char[n];
        char prev = ' ';
        b[0] = '1';
        if (a[0] == '1') {
            prev = '2';
        } else {
            prev = '1';
        }
        for (int i = 1; i < n; i++) {
            if (prev == '2') {
                if (a[i] == '1') {
                    b[i] = '0';
                    prev = '1';
                } else {
                    b[i] = '1';
                    prev = '1';
                }
            } else if (prev == '1') {
                if (a[i] == '1') {
                    b[i] = '1';
                    prev = '2';
                } else {
                    b[i] = '0';
                    prev = '0';
                }
            } else {
                if (a[i] == '1') {
                    b[i] = '1';
                    prev = '2';
                } else {
                    b[i] = '1';
                    prev = '1';
                }
            }
        }
        System.out.println(b);
    }

    static long lcm(int a, int b) {
        return (a / gcd(a, b)) * b;
    }

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

    private static long modular_add(long a, long b) {
        return ((a % mod) + (b % mod)) % mod;
    }

    private static long modular_sub(long a, long b) {
        return ((a % mod) - (b % mod) + mod) % mod;
    }

    private static long modular_mult(long a, long b) {
        return ((a % mod) * (b % mod)) % mod;
    }

    static void sort(int[] a) {
        ArrayList<Integer> l = new ArrayList<>();
        for (int i : a) l.add(i);
        Collections.sort(l);
        for (int i = 0; i < a.length; i++) a[i] = l.get(i);
    }

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

        boolean hasNext() {
            if (st != null && st.hasMoreTokens()) {
                return true;
            }
            String tmp;
            try {
                br.mark(1000);
                tmp = br.readLine();
                if (tmp == null) {
                    return false;
                }
                br.reset();
            } catch (IOException e) {
                return false;
            }
            return true;
        }

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

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

        int[] readArray(int n) {
            int[] a = new int[n];
            for (int i = 0; i < n; i++) a[i] = nextInt();
            return a;
        }

        long[] readArrayLong(int n) {
            long[] a = new long[n];
            for (int i = 0; i < n; i++) a[i] = nextLong();
            return a;
        }

        long nextLong() {

            return Long.parseLong(next());
        }

        double nextDouble() {
            return Double.parseDouble(next());
        }
    }
}