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.util.*;import java.io.*;import java.math.*;
public class FuturePuzzle
{
    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 void ruffleSort(long[] a) {
        int n = a.length; 
        for (int i=0; i<n; i++) {
            int oi=random.nextInt(n);
            long temp=a[oi];
            a[oi]=a[i]; a[i]=temp;
        }
        ArrayList<Long> lst = new ArrayList<>();
        for(long i : a)
            lst.add(i);
        Collections.sort(lst);
        for(int i = 0; i < n; i++)
            a[i] = lst.get(i);
    }
 
    public static void process()throws IOException
    {
        int n = ni();
        char[] ch = nln().toCharArray();
        int[] a = new int[n];
        for(int i = 0; i < n; i++)
            a[i] = ch[i]-'0';
        int[] b = new int[n];
        b[0] = a[0]+1;
        for(int i = 1; i < n; i++){
            if(a[i]+1 == b[i-1])
                b[i] = a[i];
            else
                b[i] = a[i]+1;
        }
        for(int i = 0; i < n; i++)
            p(b[i]-a[i]);
        pn("");
    }
   
    static AnotherReader sc;
    static PrintWriter out;
    public static void main(String[]args)throws IOException
    {
        boolean oj = System.getProperty("ONLINE_JUDGE") != null;
        if(oj){sc=new AnotherReader();out=new PrintWriter(System.out);}
        else{sc=new AnotherReader(100);out=new PrintWriter("output.txt");}
        long s = System.currentTimeMillis();
        int t=1;
        t=ni();
        //int k = t;
        while(t-->0) {/*p("Case #"+ (k-t) + ": ")*/;process();}
        out.flush();
        System.err.println(System.currentTimeMillis()-s+"ms");
        out.close();
    }

    static long power(long k, long c, long mod){
        long y = 1;
        while(c > 0){
            if(c%2 == 1)
                y = y * k % mod;
            c = c/2;
            k = k * k % mod;
        }
        return y;
    }

 
    static void pn(Object o){out.println(o);}
    static void p(Object o){out.print(o);}
    static void pni(Object o){out.println(o);out.flush();}
    static int ni()throws IOException{return sc.nextInt();}
    static long nl()throws IOException{return sc.nextLong();}
    static double nd()throws IOException{return sc.nextDouble();}
    static String nln()throws IOException{return sc.nextLine();}
    static int[] nai(int N)throws IOException{int[]A=new int[N];for(int i=0;i!=N;i++){A[i]=ni();}return A;}
    static long[] nal(int N)throws IOException{long[]A=new long[N];for(int i=0;i!=N;i++){A[i]=nl();}return A;}
    static long gcd(long a, long b)throws IOException{return (b==0)?a:gcd(b,a%b);}
    static int gcd(int a, int b)throws IOException{return (b==0)?a:gcd(b,a%b);}
    static int bit(long n)throws IOException{return (n==0)?0:(1+bit(n&(n-1)));}
    
/////////////////////////////////////////////////////////////////////////////////////////////////////////
 
    static class AnotherReader{BufferedReader br; StringTokenizer st;
    AnotherReader()throws FileNotFoundException{
    br=new BufferedReader(new InputStreamReader(System.in));}
    AnotherReader(int a)throws FileNotFoundException{
    br = new BufferedReader(new FileReader("input.txt"));}
    String next()throws IOException{
    while (st == null || !st.hasMoreElements()) {try{
    st = new StringTokenizer(br.readLine());}
    catch (IOException  e){ e.printStackTrace(); }}
    return st.nextToken(); } int nextInt() throws IOException{
    return Integer.parseInt(next());}
    long nextLong() throws IOException
    {return Long.parseLong(next());}
    double nextDouble()throws IOException { return Double.parseDouble(next()); }
    String nextLine() throws IOException{ String str = ""; try{
    str = br.readLine();} catch (IOException e){
    e.printStackTrace();} return str;}}

/////////////////////////////////////////////////////////////////////////////////////////////////////////////
}