A. Qingshan Loves Strings 2time limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard outputQingshan has a string s which only contains \texttt{0} and \texttt{1}.A string a of length k is good if and only if  a_i \ne a_{k-i+1} for all i=1,2,\ldots,k. For Div. 2 contestants, note that this condition is different from the condition in problem B.For example, \texttt{10}, \texttt{1010}, \texttt{111000} are good, while \texttt{11}, \texttt{101}, \texttt{001}, \texttt{001100} are not good.Qingshan wants to make s good. To do this, she can do the following operation at most 300 times (possibly, zero):   insert \texttt{01} to any position of s (getting a new s). Please tell Qingshan if it is possible to make s good. If it is possible, print a sequence of operations that makes s good.InputThe input consists of multiple test cases. The first line contains a single integer t (1\le t\le 100) — the number of test cases. The description of the test cases follows.The first line of each test case contains a single integer n (1 \le n\le 100) — the length of string s, respectively.The second line of each test case contains a string s with length n.It is guaranteed that s only consists of \texttt{0} and \texttt{1}.OutputFor each test case, if it impossible to make s good, output -1.Otherwise, output p (0 \le p \le 300) — the number of operations, in the first line.Then, output p integers in the second line. The i-th integer should be an index x_i (0 \le x_i \le n+2i-2) — the position where you want to insert \texttt{01} in the current s. If x_i=0, you insert \texttt{01} at the beginning of s. Otherwise, you insert \texttt{01} immediately after the x_i-th character of s.We can show that under the constraints in this problem, if an answer exists, there is always an answer that requires at most 300 operations.ExampleInput
620130004111160011101001110011003001Output
0

-1
-1
2
6 7
1
10
-1
NoteIn the first test case, you can do zero operations and get s=\texttt{01}, which is good.Another valid solution is to do one operation: (the inserted \texttt{01} is underlined)  \texttt{0}\underline{\texttt{01}}\texttt{1} and get s = \texttt{0011}, which is good.In the second and the third test case, it is impossible to make s good.In the fourth test case, you can do two operations:  \texttt{001110}\underline{\texttt{01}}  \texttt{0011100}\underline{\texttt{01}}\texttt{1} and get s = \texttt{0011100011}, which is good.
620130004111160011101001110011003001
0  -1 -1 2 6 7 1 10 -1 
['constructive algorithms', 'greedy', 'implementation', '*1300']