CF1474A.Puzzle From the Future

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.

The first line contains a single integer $t$ ( $1 \leq t \leq 1000$ ) — the number of test cases.

The first line of each test case contains the integer $n$ ( $1 \leq n \leq 10^5$ ) — 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 $10^5$ .

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 $n$ .

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 $d$ .