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 Main {
	static class FastReader
	{
	BufferedReader br;
	StringTokenizer st;

	public FastReader()
	{
	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());
	}

	String nextLine()
	{
	String str = "";
	try
	{
	str = br.readLine();
	}
	catch (IOException e)
	{
	e.printStackTrace();
	}
	return str;
	}
	}
	static int m = (int)(1e9+9);
	static long mod(long x) {
		return ((x%m + m)%m);
	}
	static long add(long a,long b) {
		return mod(mod(a)+mod(b));
	}
	static long mul(long a,long b) {
		return mod(mod(a) * mod(b));
	}
	
	public static void main(String[] args) {
		FastReader sc = new FastReader();
		int t = 1;
		t = sc.nextInt();
		
		while(t-->0) {
			int n = sc.nextInt();
			String b = sc.next();
			String a = "1";
			for(int i=1;i<b.length();i++) {
				int x = Integer.parseInt(a.charAt(i-1)+"") + Integer.parseInt(b.charAt(i-1)+"");
				int y = Integer.parseInt(b.charAt(i)+"");
				if(x==2) {
					if(y==0)
					a = a + "1";
					else a = a+"0";
				}
				else if(x==1) {
					if(y==1)
						a = a+"1";
					if(y==0)
						a = a + "0";
				}
				else {
					if(y==1) a = a+"1";
					else a = a+"1";
				}
			}
			System.out.println(a);
		}
	}
	private static long lb(ArrayList<Long> list, long l) {
		int lo = 0,h = list.size()-1;
		int ans = -1;
		while(lo<=h) {
			int mid = lo+(h-lo)/2;
			if(list.get(mid)<=l) {
				if(mid>0 && list.get(mid-1)==l) h = mid - 1;
				else {
					lo = mid+1;
					ans = mid;
				}
			}
			else {
				h = mid-1;
			}
		}
		return ans+1;
	}
	
	
	private static int ub(List<Long> list, long l) {
		int lo = 0,h = list.size()-1;
		int ans = -1;
		while(lo<=h) {
			int mid = lo+(h-lo)/2;
			if(list.get(mid)<=l) {
				lo = mid+1;
				ans = mid;
			}
			else {
				
				h = mid-1;
			}
		}
		
		return ans+1;
	}
	private static int check(int i, int x) {
		int cnt = 0;
		while(i%x==0 && i>0) {
			i = i/x;
			cnt++;
		}
		return cnt;
	}
	static String reverse(String x) {
		StringBuilder input1 = new StringBuilder();
		
        input1.append(x);
        
        input1 = input1.reverse();
        
        return input1.toString();
	}
	private static long lsbt(long val) {
		long msk = val&(val-1);
		return (int) (Math.log(val^msk)/Math.log(2));
	}
	private static int sqrt(long x) {
		long l = 1,h = x;
		int ans = 0;
		while(l<=h) {
			long mid = (l+(h-l)/2);
			if(mid>x/mid) h = mid-1;
			else {
				ans = (int) mid;
				l = mid+1;
			}
		}
		return ans;
	}
}
class pair implements Comparable<pair>{
	int a;
	int b;
	pair(int a,int b){
		this.a = a;
		this.b = b;
	}
	@Override
	public int compareTo(pair o) {
		// TODO Auto-generated method stub
		return o.b - this.b;
	}
}
class spair{
	pair p;
	int id;
	spair(pair p,int id){
		this.p = p;
		this.id = id;
	}
}