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 {

	static class pair implements Comparable<pair> {

		int f;
		int s;

		public pair() {

		}

		public pair(int a, int b) {
			f = a;
			s = b;

		}

		@Override
		public int compareTo(pair o) {
			// TODO Auto-generated method stub
			return o.s - this.s;
		}

	}

	static int mod = (int) 1e9 + 7;
	static int ar[];
	static Scanner sc = new Scanner(System.in);
	static StringBuilder out = new StringBuilder();
	static ArrayList<Integer> gr[]=new ArrayList[100000];
	

	static void buildGraph(int n, int m) throws IOException {

		gr = new ArrayList[n];

		for (int i = 0; i < n; i++) {

			gr[i] = new ArrayList<>();

		}

		for (int i = 0; i < m; i++) {
			int u = sc.nextInt();
			int v = sc.nextInt();

			u--;
			v--;
			gr[u].add(v);
			gr[v].add(u);
		}

	}

	static void sort(int a[], int n) {

		ArrayList<Integer> al = new ArrayList<>();

		for (int i = 0; i < n; i++) {

			al.add(a[i]);
		}

		Collections.sort(al);

		for (int i = 0; i < n; i++) {

			a[i] = al.get(i);
		}
	}

	static void in(int a[], int n) throws IOException {
		for (int i = 0; i < n; i++)
			a[i] = sc.nextInt();
	}

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

		int t = sc.nextInt();
		while (t-- > 0) {

			Math.pow(2, 64);
			int cnt=0,id=0,result=0,ans=0;
			int n = sc.nextInt();
			String s = sc.next();
			char ch[] = s.toCharArray();
			boolean ok = false;

			out.append(1);

			int check = (ch[0] == '1' ? 2 : 1);
			for (int i = 1; i < n; i++) {

				if (check == 0) {

					out.append(1);

					if (ch[i] == '1')
						check = 2;
					else
						check = 1;
				} else if (check == 1) {

					if (ch[i] == '1') {

						out.append(1);
						check = 2;

					} else {

						out.append(0);
						check = 0;

					}
				} else {

					if (ch[i] == '0')
						out.append(1);
					else
						out.append(0);
					check = 1;
				}
			}
			out.append("\n");

		}
		System.out.println(out);

	}

}
