CF1791D.Distinct Split

Let's denote the $f(x)$ function for a string $x$ as the number of distinct characters that the string contains. For example $f(\texttt{abc}) = 3$ , $f(\texttt{bbbbb}) = 1$ , and $f(\texttt{babacaba}) = 3$ .

Given a string $s$ , split it into two non-empty strings $a$ and $b$ such that $f(a) + f(b)$ is the maximum possible. In other words, find the maximum possible value of $f(a) + f(b)$ such that $a + b = s$ (the concatenation of string $a$ and string $b$ is equal to string $s$ ).

The input consists of multiple test cases. The first line contains an integer $t$ ( $1 \leq t \leq 10^4$ ) — the number of test cases. The description of the test cases follows.

The first line of each test case contains an integer $n$ ( $2 \leq n \leq 2\cdot10^5$ ) — the length of the string $s$ .

The second line contains the string $s$ , consisting of lowercase English letters.

It is guaranteed that the sum of $n$ over all test cases does not exceed $2\cdot10^5$ .

For each test case, output a single integer — the maximum possible value of $f(a) + f(b)$ such that $a + b = s$ .

For the first test case, there is only one valid way to split $\texttt{aa}$ into two non-empty strings $\texttt{a}$ and $\texttt{a}$ , and $f(\texttt{a}) + f(\texttt{a}) = 1 + 1 = 2$ .

For the second test case, by splitting $\texttt{abcabcd}$ into $\texttt{abc}$ and $\texttt{abcd}$ we can get the answer of $f(\texttt{abc}) + f(\texttt{abcd}) = 3 + 4 = 7$ which is maximum possible.

For the third test case, it doesn't matter how we split the string, the answer will always be $2$ .