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.
#include <iostream>
#include <vector> 
using namespace std;

int main()
{long int t;
cin>>t;
while(t>0)
{
    

    long int i,j,k,l,p,q,n;
    
    string s;
    vector<long int>a;
       cin>>n;
 long int v[n+1];
 
    cin>>s;

    v[0]=1;
  // cout<<1;
    for(i=1;i<n;i++)
    {
        if(s[i]=='1'&&s[i-1]=='1')
        {//cout<<'o';
            if(v[i-1]==1)
            {
                v[i]=0;
            }
            else
            {
                v[i]=1;
            }
    
    }
 else  if(s[i]=='0'&&s[i-1]=='0')
        {//cout<<'p';
            if(v[i-1]==1)
            {
                v[i]=0;
            }
            else
            {
                v[i]=1;
            }
    
    }
    
    else
    {//cout<<'h';
          if(s[i]=='0')
            {//cout<<'h';
                if(v[i-1]==1)
                {
                    v[i]=1;
                }
                else
                {//cout<<'q';
                    v[i]=0;
                }
            }
            else
            {
              v[i]=1;
            }
    }
    
    
    }
    for(i=0;i<n;i++)
    {
        cout<<v[i];
    }
    cout<<'\n';
    t--;
} 

}