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.*;
import java.util.*;

public class A {


	public static void main(String[] args)throws IOException {

		InputReader in = new InputReader();
		PrintWriter pw = new PrintWriter(System.out);
		int t = in.nextInt();
		while (t-- > 0) {
			int n = in.nextInt();
			char []a = in.nextLine().toCharArray();
			int last = -1;
			char []b = new char[n];

			for (int i = 0; i < a.length; i++) {
				if (a[i] == '0') {
					if (last == 1) {
						b[i] = '0';
						last = 0;
					} else {
						b[i] = '1';
						last = 1;
					}
				} else {
					if (last == 2) {
						b[i] = '0';
						last = 1;
					} else {
						b[i] = '1';
						last = 2;
					}
				}
			}
			pw.println(String.valueOf(b));
		}

		pw.close();
	}

	static final Random random = new Random();

	static void ruffleSort(int[] a) {
		int n = a.length;
		for (int i = 0; i < n; i++) {
			int oi = random.nextInt(n), temp = a[oi];
			a[oi] = a[i];
			a[i] = temp;
		}
		ArrayList<Integer> lst = new ArrayList<>();
		for (int i : a)
			lst.add(i);
		Collections.sort(lst);
		for (int i = 0; i < n; i++)
			a[i] = lst.get(i);
	}

	static class InputReader {
		BufferedReader br;
		StringTokenizer st;

		public InputReader() {
			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());
		}
		int[] readArray(int n) {
			int []a = new int[n];
			for (int i = 0; i < n; i++)
				a[i] = nextInt();
			return a;
		}
		long[] readLongArray(int n) {
			long []a = new long[n];
			for (int i = 0; i < n; i++)
				a[i] = nextLong();
			return a;
		}
		int [][] read2dArray(int n, int m) {
			int [][]a = new int[n][m];
			for (int i = 0; i < n; i++)
				for (int j = 0; j < m; j++)
					a[i][j] = nextInt();
			return a;
		}
		String nextLine() {
			String str = "";
			try {
				str = br.readLine();
			} catch (IOException e) {
				e.printStackTrace();
			}
			return str;
		}
	}
}