CF2178A Yes or Yes

Last Christmas, your friend Fernando gifted you a string $ s $ consisting only of the characters $ \mathtt{Y} $ and $ \mathtt{N} $ , representing "Yes" and "No", respectively. You can repeatedly apply the following operation on $ s $ : - Choose any two adjacent characters and replace them with their [logical OR](https://en.wikipedia.org/wiki/Logical_disjunction). Formally, in each operation, you can choose an index $ i $ ( $ 1 \leq i \leq |s|-1 $ ), remove the characters $ s_i $ and $ s_{i+1} $ , then insert: - A single $ \mathtt{Y} $ if at least one of $ s_i $ or $ s_{i+1} $ is $ \mathtt{Y} $ ; - A single $ \mathtt{N} $ if both $ s_i $ and $ s_{i+1} $ are $ \mathtt{N} $ . Note that after each operation, the length of $ s $ decreases by $ 1 $ . Unfortunately, Fernando does not want you to combine "Yes OR Yes", as he has experienced trauma relating to a certain song. Determine whether it is possible to reduce $ s $ to a single character by repeatedly applying the operation above, without ever combining two $ \mathtt{Y} $ 's.

Each test contains multiple test cases. The first line contains the number of test cases $ t $ ( $ 1 \le t \le 500 $ ). The description of the test cases follows. The only line of each test case contains the string $ s $ ( $ 2\le |s|\le 100 $ ). It is guaranteed that $ s_i = \mathtt{Y} $ or $ \mathtt{N} $ .

For each test case, print "YES" if the string can be reduced to a single character by repeatedly applying the described operation, and "NO" otherwise. You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

In the first test case, you cannot combine $ s_1 $ and $ s_2 $ since they are both $ \mathtt{Y} $ . Thus, the answer is NO. In the third test case, the following is a valid sequence of operations: $ \mathtt{\underline{NN}Y}\to\mathtt{\underline{NY}}\to\mathtt{Y} $ . Thus, the answer is YES. In the fourth test case, there are two possibilities for the first operation: $ \mathtt{YY\underline{YN}Y}\to \mathtt{YYYY} $ or $ \mathtt{YYY\underline{NY}}\to \mathtt{YYYY} $ . However, in either case, it is not possible to perform any more operations without combining two $ \mathtt{Y} $ 's. Thus, the answer is NO. In the fifth test case, the following is a valid sequence of operations: $ \mathtt{N\underline{NN}NN}\to\mathtt{\underline{NN}NN}\to\mathtt{N\underline{NN}}\to\mathtt{\underline{NN}}\to\mathtt{N} $ . Thus, the answer is YES.