This is an interactive problem.

For an $$$n$$$-digit ternary number, Little A designed a very simple encryption algorithm: for each digit $$$i=0,1,\cdots,n-1$$$, construct a bijection $$$f_i:\{0,1,2\}\to\{0,1,2\}$$$. For an $$$n$$$-digit ternary number $$$A$$$, suppose its digits from the most significant to the least significant are: $$$$$$ a_{n-1},a_{n-2},\cdots,a_{0} $$$$$$

Then its encrypted form from the most significant to the least significant is: $$$$$$ f_{n-1}(a_{n-1}),f_{n-2}(a_{n-2}),\cdots,f_0(a_0) $$$$$$

To conveniently represent the encryption of each digit, we can denote the bijection $$$f_i=\{0\to x,1\to y,2\to z\}$$$ (where $$$x,y,z\in\{0,1,2\}$$$ and $$$x,y,z$$$ are distinct) simply as $$$f_i=xyz$$$. For example, $$$f_0=\{0\to 1,1\to 2,2\to 0\}$$$ can be represented as $$$f_0=120$$$.

For instance, if Little A constructs the bijections $$$f_0=021,f_1=120,f_2=210$$$ for a 3-digit ternary number, then the encryption of $$$120$$$ results in $$$100$$$, and the encryption of $$$201$$$ results in $$$012$$$.

Of course, Little A does not want others to know the bijections he constructed, but the encryption algorithm must be used, so Little A will tell you the size of $$$n$$$ and allow you to inquire about the results of addition in the encrypted state twice. Your task is to use these 2 inquiries to decipher the $$$n$$$ bijections $$$f_0,f_1,\cdots,f_{n-1}$$$ constructed by Little A.

You can make at most 2 inquiries, after which you must answer Little A's encryption mappings.

For each inquiry, you should provide the two parameters $$$a,b$$$ for addition in the format ? $$$a$$$ $$$b$$$, where $$$a,b$$$ are both $$$n$$$-digit ternary numbers and must be provided in string format. After flushing the output stream, you need to read a line containing an $$$n+1$$$-digit ternary number $$$c$$$ in string format, where $$$c$$$ is the result of adding the decrypted forms of $$$a$$$ and $$$b$$$, followed by encryption, with the most significant digit serving as an overflow flag that is not encrypted. Formally: $$$$$$ c=f(f^{-1}(a)+f^{-1}(b)) $$$$$$

where $$$f$$$ is the encryption bijection, and $$$f^{-1}$$$ is the inverse mapping of $$$f$$$.

Note that all ternary strings are input from the most significant to the least significant digit!

Note that the highest bit (the $$$n$$$-th bit) of $$$c$$$ is treated as an overflow flag and is not encrypted, or you may consider $$$f_n = 012$$$.

To answer Little A's encryption mapping, you should provide the $$$n$$$ encryption mappings in the format ! $$$f_0$$$ $$$f_1$$$ $$$\cdots$$$ $$$f_{n-1}$$$, where each $$$f_0,f_1,\cdots,f_{n-1}$$$ should be one of the strings 012, 021, 102, 120, 201, 210. After flushing the output stream, the interactor will immediately determine whether your answer is correct, and then proceed to the next test case or end the program without any extra output.

Note that Little A's encryption mappings are fully determined before the inquiries and will not change with the inquiries. In other words, the interactor is not adaptive.

To flush the output stream, you can:

• Use fflush(stdout) (if using printf) or cout.flush() (if using cout) in C/C++.
• Use System.out.flush() in Java.
• Use sys.stdout.flush() in Python.