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{

	//public static Reader sc = new Reader();
	public static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

   public static void main(String[] args) throws IOException{
	
	int t = Integer.parseInt(br.readLine());
	while (t-->0){
		int n = Integer.parseInt(br.readLine());
		String s = br.readLine();
		int prev = -1;
		for (int i = 0; i < n; i++) {
			if (s.charAt(i) == '1') {
				if (prev != 2) {
					out.print('1');
					prev = 2;
				}
				else {
					out.print('0');
					prev = 1;
				}
			}
			else {
				if (prev != 1) {
					out.print('1');
					prev = 1;
				}
				else {
					out.print('0');
					prev = 0;
				}
			}
		}
		out.println();
	}
	
	out.close();
   }
   static long ceil(long a, long b) {
		return (a+b-1)/b;
	}
  
   static long powMod(long base, long exp, long mod) {
	   if (base == 0 || base == 1) return base;
	   if (exp == 0) return 1;
	   if (exp == 1) return base % mod;
	   long R = powMod(base, exp/2, mod) % mod;
	   R *= R;
	   R %= mod;
	   if ((exp & 1) == 1) {
		   return base * R % mod;
	   }
	   else return R % mod;
   }
   
   static long pow(long base, long exp) {
	   if (base == 0 || base == 1) return base;
	   if (exp == 0) return 1;
	   if (exp == 1) return base;
	   long R = pow(base, exp/2);
	   if ((exp & 1) == 1) {
		   return R * R * base;
	   }
	   else return R * R;
   }
   

 
   static class Reader 
   { 
       final private int BUFFER_SIZE = 1 << 16; 
       private DataInputStream din; 
       private byte[] buffer; 
       private int bufferPointer, bytesRead; 

       public Reader() 
       { 
           din = new DataInputStream(System.in); 
           buffer = new byte[BUFFER_SIZE]; 
           bufferPointer = bytesRead = 0; 
       } 

       public Reader(String file_name) throws IOException 
       { 
           din = new DataInputStream(new FileInputStream(file_name)); 
           buffer = new byte[BUFFER_SIZE]; 
           bufferPointer = bytesRead = 0; 
       } 

       public String readLine() throws IOException 
       { 
           byte[] buf = new byte[64]; // line length 
           int cnt = 0, c;
           while ((c = read()) != -1) 
           { 
               if (c == '\n') 
               {
            	   break;
               }
               buf[cnt++] = (byte) c; 
           } 
           return new String(buf, 0, Integer.max(cnt-1, 0)); 
       } 

       public int nextInt() throws IOException 
       { 
           int ret = 0; 
           byte c = read(); 
           while (c <= ' ') 
               c = read(); 
           boolean neg = (c == '-'); 
           if (neg) 
               c = read(); 
           do
           { 
               ret = ret * 10 + c - '0'; 
           }  while ((c = read()) >= '0' && c <= '9'); 

           if (neg) 
               return -ret; 
           return ret; 
       } 

       public long nextLong() throws IOException 
       { 
           long ret = 0; 
           byte c = read(); 
           while (c <= ' ') 
               c = read(); 
           boolean neg = (c == '-'); 
           if (neg) 
               c = read(); 
           do { 
               ret = ret * 10 + c - '0'; 
           } 
           while ((c = read()) >= '0' && c <= '9'); 
           if (neg) 
               return -ret; 
           return ret; 
       } 

       public double nextDouble() throws IOException 
       { 
           double ret = 0, div = 1; 
           byte c = read(); 
           while (c <= ' ') 
               c = read(); 
           boolean neg = (c == '-'); 
           if (neg) 
               c = read(); 

           do { 
               ret = ret * 10 + c - '0'; 
           } 
           while ((c = read()) >= '0' && c <= '9'); 

           if (c == '.') 
           { 
               while ((c = read()) >= '0' && c <= '9') 
               { 
                   ret += (c - '0') / (div *= 10); 
               } 
           } 

           if (neg) 
               return -ret; 
           return ret; 
       } 

       private void fillBuffer() throws IOException 
       { 
           bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE); 
           if (bytesRead == -1) 
               buffer[0] = -1; 
       } 

       private byte read() throws IOException 
       { 
           if (bufferPointer == bytesRead) 
               fillBuffer(); 
           return buffer[bufferPointer++]; 
       } 

       public void close() throws IOException 
       { 
           if (din == null) 
               return; 
           din.close(); 
       } 
   }

	public static PrintWriter out = new PrintWriter (new BufferedOutputStream(System.out));
}











