Typical Interview Problem
The FB-string is formed as follows. Initially, it is empty. We go through all positive integers, starting from $$$1$$$, in ascending order, and do the following for each integer:  if the current integer is divisible by $$$3$$$, append F to the end of the FB-string;  if the current integer is divisible by $$$5$$$, append B to the end of the FB-string. Note that if an integer is divisible by both $$$3$$$ and $$$5$$$, we append F, and then B, not in the opposite order.The first $$$10$$$ characters of the FB-string are FBFFBFFBFB: the first F comes from the integer $$$3$$$, the next character (B) comes from $$$5$$$, the next F comes from the integer $$$6$$$, and so on. It's easy to see that this string is infinitely long. Let $$$f_i$$$ be the $$$i$$$-th character of FB-string; so, $$$f_1$$$ is F, $$$f_2$$$ is B, $$$f_3$$$ is F, $$$f_4$$$ is F, and so on.You are given a string $$$s$$$, consisting of characters F and/or B. You have to determine whether it is a substring (contiguous subsequence) of the FB-string. In other words, determine if it is possible to choose two integers $$$l$$$ and $$$r$$$ ($$$1 \le l \le r$$$) so that the string $$$f_l f_{l+1} f_{l+2} \dots f_r$$$ is exactly $$$s$$$.For example:  FFB is a substring of the FB-string: if we pick $$$l = 3$$$ and $$$r = 5$$$, the string $$$f_3 f_4 f_5$$$ is exactly FFB;  BFFBFFBF is a substring of the FB-string: if we pick $$$l = 2$$$ and $$$r = 9$$$, the string $$$f_2 f_3 f_4 \dots f_9$$$ is exactly BFFBFFBF;  BBB is not a substring of the FB-string. 
The first line contains one integer $$$t$$$ ($$$1 \le t \le 2046$$$) — the number of test cases.Each test case consists of two lines. The first line contains one integer $$$k$$$ ($$$1 \le k \le 10$$$) — the number of characters in $$$s$$$. The second line contains $$$s$$$, which is a string of exactly $$$k$$$ characters. Each character of $$$s$$$ is either F or B.
For each test case, print YES if $$$s$$$ is a substring of the FB-string, or NO otherwise.You may print each letter in any case (YES, yes, Yes will all be recognized as positive answer, NO, no and nO will all be recognized as negative answer).
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tp = 'FBFFBFFBFBFFBFFBFB'

def i():
    n = int(input())
    s = input()
    if s in tp:
        return 'YES\n'
    return 'NO\n'
t = int(input())
a = []
for _ in range(t):
    a.append(i())
print(''.join(a), end='')
for s in [*open(0)][2::2]:
    print('NYOE S'[s[:-1] in 'FBFFBFFB' * 3::2])
for s in [*open(0)][2::2]:
    print('NYOE S'[s[:-1] in 'FBFFBFFB' * 3::2])