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 $s$).

我们用 $f(x)$ 表示字符串 $x$ 的不同字符个数。例如：$f(\texttt{abc}) = 3$，$f(\texttt{bbbbb}) = 1$，$f(\texttt{babacaba}) = 3$。

给定一个字符串 $s$，将其分割为两个非空字符串 $a$ 和 $b$，使得 $f(a) + f(b)$ 尽可能大。换言之，求满足 $a + b = s$（即字符串 $a$ 与字符串 $b$ 的拼接等于字符串 $s$）时，$f(a) + f(b)$ 的最大可能值。

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 $2\cdot10^5$.

输入包含多个测试用例。第一行包含一个整数 $t$（$1 \leq t \leq 10^4$），表示测试用例的数量。随后是各测试用例的描述。

每个测试用例的第一行包含一个整数 $n$（$2 \leq n \leq 2\cdot10^5$），表示字符串 $s$ 的长度。

每个测试用例的第二行包含字符串 $s$，由小写英文字母组成。

保证所有测试用例的 $n$ 之和不超过 $2\cdot10^5$。

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

对于每个测试用例，输出一个整数——满足 $a + b = s$ 的 $f(a) + f(b)$ 的最大可能值。

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 $2$.

对于第一个测试用例，将字符串 $\texttt{aa}$ 拆分为两个非空字符串 $\texttt{a}$ 和 $\texttt{a}$ 只有一种合法方式，且 $f(\texttt{a}) + f(\texttt{a}) = 1 + 1 = 2$。

对于第二个测试用例，将字符串 $\texttt{abcabcd}$ 拆分为 $\texttt{abc}$ 和 $\texttt{abcd}$，可得到 $f(\texttt{abc}) + f(\texttt{abcd}) = 3 + 4 = 7$，这是可能的最大值。

对于第三个测试用例，无论以何种方式拆分该字符串，答案恒为 $2$。