Two Subsequences
You are given a string $$$s$$$. You need to find two non-empty strings $$$a$$$ and $$$b$$$ such that the following conditions are satisfied:  Strings $$$a$$$ and $$$b$$$ are both subsequences of $$$s$$$.  For each index $$$i$$$, character $$$s_i$$$ of string $$$s$$$ must belong to exactly one of strings $$$a$$$ or $$$b$$$.  String $$$a$$$ is lexicographically minimum possible; string $$$b$$$ may be any possible string. Given string $$$s$$$, print any valid $$$a$$$ and $$$b$$$.Reminder:A string $$$a$$$ ($$$b$$$) is a subsequence of a string $$$s$$$ if $$$a$$$ ($$$b$$$) can be obtained from $$$s$$$ by deletion of several (possibly, zero) elements. For example, "dores", "cf", and "for" are subsequences of "codeforces", while "decor" and "fork" are not.A string $$$x$$$ is lexicographically smaller than a string $$$y$$$ if and only if one of the following holds:  $$$x$$$ is a prefix of $$$y$$$, but $$$x \ne y$$$;  in the first position where $$$x$$$ and $$$y$$$ differ, the string $$$x$$$ has a letter that appears earlier in the alphabet than the corresponding letter in $$$y$$$. 
In the first test case, there are only two choices: either $$$a =$$$ f and $$$b = $$$ c or $$$a = $$$ c and $$$b = $$$ f. And $$$a = $$$c is lexicographically smaller than $$$a = $$$ f.In the second test case, a is the only character in the string.In the third test case, it can be proven that b is the lexicographically smallest subsequence of $$$s$$$. The second string can be of two variants; one of them is given here.
Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 1000$$$). Description of the test cases follows.The first and only line of each test case contains one string $$$s$$$ ($$$2 \le |s| \le 100$$$ where $$$|s|$$$ means the length of $$$s$$$). String $$$s$$$ consists of lowercase Latin letters.
For each test case, print the strings $$$a$$$ and $$$b$$$ that satisfy the given conditions. If there are multiple answers, print any.
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def i():
    s = input()
    c = s.index(min(s))
    return f'{s[c]} {s[:c] + s[c + 1:]}'
t = int(input())
a = []
for _ in range(t):
    a.append(i())
print('\n'.join(a))
for s in [*open(0)][1:]:
    i = s.find((m := min(s[:-1])))
    print(m, s[:i] + s[i + 1:])
for s in [*open(0)][1:]:
    i = s.find((m := min(s[:-1])))
    print(m, s[:i] + s[i + 1:])