CF1919B.Plus-Minus Split

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题目描述

You are given a string $s$ of length $n$ consisting of characters "+" and "-". $s$ represents an array $a$ of length $n$ defined by $a_i=1$ if $s_i=$ "+" and $a_i=-1$ if $s_i=$ "-".

You will do the following process to calculate your penalty:

Split $a$ into non-empty arrays $b_1,b_2,\ldots,b_k$ such that $b_1+b_2+\ldots+b_k=a^\dagger$ , where $+$ denotes array concatenation.
The penalty of a single array is the absolute value of its sum multiplied by its length. In other words, for some array $c$ of length $m$ , its penalty is calculated as $p(c)=|c_1+c_2+\ldots+c_m| \cdot m$ .
The total penalty that you will receive is $p(b_1)+p(b_2)+\ldots+p(b_k)$ .

If you perform the above process optimally, find the minimum possible penalty you will receive.

$^\dagger$ Some valid ways to split $a=[3,1,4,1,5]$ into $(b_1,b_2,\ldots,b_k)$ are $([3],[1],[4],[1],[5])$ , $([3,1],[4,1,5])$ and $([3,1,4,1,5])$ while some invalid ways to split $a$ are $([3,1],[1,5])$ , $([3],[\,],[1,4],[1,5])$ and $([3,4],[5,1,1])$ .

输入格式

Each test contains multiple test cases. The first line contains a single integer $t$ ( $1 \leq t \leq 1000$ ) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer $n$ ( $1 \le n \le 5000$ ) — the length of string $s$ .

The second line of each test case contains string $s$ ( $s_i \in \{ \mathtt{+}, \mathtt{-} \}$ , $|s| = n$ ).

Note that there are no constraints on the sum of $n$ over all test cases.

输出格式

For each test case, output a single integer representing the minimum possible penalty you will receive.

输入输出样例

输入#1

5
1
+
5
-----
6
+-+-+-
10
--+++++++-
20
+---++++-+++++---++-

输出#1

说明/提示

In the first test case, we have $a=[1]$ . We can split array $a$ into $([1])$ . Then, the sum of penalties of the subarrays is $p([1]) = 1$ .

In the second test case, we have $a=[-1,-1,-1,-1,-1]$ . We can split array $a$ into $([-1],[-1],[-1],[-1],[-1])$ . Then, the sum of penalties of the subarrays is $p([-1]) + p([-1]) + p([-1]) + p([-1]) + p([-1]) = 1 + 1 + 1 + 1 + 1 = 5$ .

In the third test case, we have $a=[1,-1,1,-1,1,-1]$ . We can split array $a$ into $([1,-1,1,-1],[1,-1])$ . Then, the sum of penalties of the subarrays is $p([1,-1,1,-1]) + p([1,-1]) = 0 + 0 = 0$ .

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