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.
//SHIVAM GARG
import java.util.*;
import java.lang.*;
import java.io.*;


public class Codechef
{
 public static FastReader reader = new FastReader();
 
 public static void main (String[] args) throws java.lang.Exception {
	    
	    int tests = reader.nextInt();
	    
        for(int ii=0; ii< tests; ii++){
            int n = reader.nextInt();
            int[] arr = new int[n+1];
            char[] chrs = reader.nextLine().toCharArray();
            int[] inp = new int[n+1];
            for(int i=0; i< chrs.length; i++){
                if(chrs[i] == '1'){
                    inp[i+1] = 1;
                }
            }
            
            int last;
            StringBuilder builder = new StringBuilder("");
            int temp =-1;
            for(int i=1; i<=n; i++ ){
                last = arr[i-1] + inp[i-1];
                if(inp[i] == 0){
                    if(last == 1){
                        temp = 0;
                    }else{
                        temp = 1;
                    }
                }else if(inp[i] == 1){
                    if(last == 2){
                        temp = 0;
                    }else{
                        temp = 1;
                    }
                }
                arr[i] = temp;
                // System.out.println("appending, last = " + last + ", inp-curr = " +inp[i-1] + ", temp = " + temp );
                builder.append(temp);
            }
            System.out.println(builder.toString());
            
        }
    }
  
    
   
   
    
    static int[] primes;
    
    public static void sieve(){
        primes = new int[100002]; // 0 means prime and 1 means Not prime
        int temp = 2;
        while(temp < primes.length ){
            primes[temp] = 1;
            temp = temp+2;
        }
        primes[2] = 0;
        
        for(int i = 3; i < primes.length; i = i+2){
            if(primes[i] == 0 ){
                temp = i+i;
                while(temp < primes.length ){
                    primes[temp] = 1;
                    temp = temp+i;
                }
            }
        }
    }
    
 
 
 
	
    
    
    public static HashMap<Integer, Integer> getFactors(long n){
        long ori = n;
        
        HashMap<Integer, Integer>  factors = new HashMap<Integer, Integer>();
        int temp =0;
        while(n % 2 == 0 ){
            temp++;
            n = n/2;
        }
        if(temp >0){
            factors.put(2, temp);
            // System.out.println( ori+" ==> " + 2 +" :: " + temp);
        }
        
        long sqrt = (long) Math.sqrt(n);
        for(int i= 3;(i <= sqrt) && i<=n ; i = i+2){
            if(primes[i] == 0 ){
                temp = 0;
                while(n % i == 0 ){
                    temp++;
                    n = n/i;
                }
                if(temp > 0){
                    // System.out.println( ori+" ==> " + i +" :: " + temp);
                    factors.put(i, temp);    
                }
                
            }
        }
        return factors;
        
    }
   
    
 
 /* Iterative Function to calculate  (x^y)%p in O(log y) */
 
    public static long prime = 1000000007;
    public static long[] fact;
    static long power(long x, long y) 
    { 
        // Initialize result 
        long res = 1; 
        // Update x if it is more than or 
        // equal to p 
        x = x % prime; 
                      
        while (y > 0) 
        { 
            // If y is odd, multiply x  with result 
            if (y % 2 == 1) 
                res = (res * x) % prime; 
            // y must be even now 
            y = y >> 1; // y = y/2 
            x = (x * x) % prime; 
        } 
          
        return res; 
    } 
      
    // Returns n^(-1) mod p 
    static long modInverse(long n) 
    {   return power(n, prime-2);   } 
      
    // Returns nCr % p using Fermat's little theorem. 
    static long nCrModPFermat(int n, int r) 
    { 
        // Base case 
        if (r == 0) 
            return 1; 
        long a1 = fact[n];
        long a2 = modInverse(fact[r]) % prime;
        long a3 = modInverse(fact[n-r]) % prime;
        return ((fact[n] * a2)%prime * a3) % prime; 
    } 
	
	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());  } 
        
        int[] nextIntArray(int n){
            int[] arr = new int[n];
            for(int ii=0; ii<n; ii++){
                arr[ii] = this.nextInt();
            }
            return arr;
        }
        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; 
        } 
    } 
	
	
}