B. Qingshan Loves Stringstime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard outputQingshan has a string s, while Daniel has a string t. Both strings only contain \texttt{0} and \texttt{1}.A string a of length k is good if and only if  a_i \ne a_{i+1} for all i=1,2,\ldots,k-1. For example, \texttt{1}, \texttt{101}, \texttt{0101} are good, while \texttt{11}, \texttt{1001}, \texttt{001100} are not good.Qingshan wants to make s good. To do this, she can do the following operation any number of times (possibly, zero):  insert t to any position of s (getting a new s). Please tell Qingshan if it is possible to make s good.InputThe input consists of multiple test cases. The first line contains a single integer T (1\le T\le 2000) — the number of test cases. The description of the test cases follows.The first line of each test case contains two integers n and m (1 \le n,m \le 50) — the length of the strings s and t, respectively.The second line of each test case contains a string s of length n.The third line of each test case contains a string t of length m.It is guaranteed that s and t only contain \texttt{0} and \texttt{1}.OutputFor each test case, print "YES" (without quotes), if it is possible to make s good, and "NO" (without quotes) otherwise.You can print letters in any case (upper or lower).ExampleInput
51 1103 31110103 2111006 7101100101010110 2100100100010Output
Yes
Yes
No
No
No
NoteIn the first test case, s is good initially, so you can get a good s by doing zero operations.In the second test case, you can do the following two operations (the inserted string t is underlined):  \texttt{1}\underline{\texttt{010}}\texttt{11}  \texttt{10101}\underline{\texttt{010}}\texttt{1} and get s = \texttt{101010101}, which is good.In the third test case, there is no way to make s good after any number of operations.
51 1103 31110103 2111006 7101100101010110 2100100100010
Yes Yes No No No 
['constructive algorithms', 'implementation', '*800']