3 ba 3 abc 4 cbaa 5
7 We have `ba`, `abc`, and `cbaa`. For example, we can use `ba` once and `abc` once for a cost of $7$, then concatenate them in the order `abc`, `ba` to make a palindrome. Also, we can use `abc` once and `cbaa` once for a cost of $9$, then concatenate them in the order `cbaa`, `abc` to make a palindrome. We cannot make a palindrome for a cost less than $7$, so we should print $7$.
2 abcab 5 cba 3
11 We can choose `abcab` once and `cba` twice, then concatenate them in the order `abcab`, `cba`, `cba` to make a palindrome for a cost of $11$.
4 ab 5 cba 3 a 12 ab 10
8 We can choose `a` once, which is already a palindrome, but it is cheaper to concatenate `ab` and `cba`.
2 abc 1 ab 2
\-1 We cannot make a palindrome, so we should print $-1$.
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