A. Puzzle From the Futuretime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard outputIn 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.InputThe 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.OutputFor 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.ExampleInput
5
1
0
3
011
3
110
6
111000
6
001011
Output
1
110
100
101101
101110
NoteIn 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.
5
1
0
3
011
3
110
6
111000
6
001011

1 110 100 101101 101110 
['greedy', '*800']