CF1750E.Bracket Cost

Daemon Targaryen decided to stop looking like a Metin2 character. He turned himself into the most beautiful thing, a bracket sequence.

For a bracket sequence, we can do two kind of operations:

• Select one of its substrings $^\dagger$ and cyclic shift it to the right. For example, after a cyclic shift to the right, "(())" will become ")(()";
• Insert any bracket, opening '(' or closing ')', wherever you want in the sequence.

We define the cost of a bracket sequence as the minimum number of such operations to make it balanced $^\ddagger$ .

Given a bracket sequence $s$ of length $n$ , find the sum of costs across all its $\frac{n(n+1)}{2}$ non-empty substrings. Note that for each substring we calculate the cost independently.

$^\dagger$ A string $a$ is a substring of a string $b$ if $a$ can be obtained from $b$ by deletion of several (possibly, zero or all) characters from the beginning and several (possibly, zero or all) characters from the end.

$^\ddagger$ A sequence of brackets is called balanced if one can turn it into a valid math expression by adding characters $+$ and $1$ . For example, sequences "(())()", "()", and "(()(()))" are balanced, while ")(", "(()", and "(()))(" are not.

Each test consists of multiple test cases. The first line contains a single integer $t$ ( $1 \leq t \leq 10^5$ ) — the number of test cases. The description of test cases follows.

The first line of each test case contains a single integer $n$ ( $1 \le n \le 2 \cdot 10^5$ ) — the length of the bracket sequence.

The second line of each test case contains a string $s$ , consisting only of characters '(' and ')', of length $n$ — the bracket sequence.

It is guaranteed that sum of $n$ across all test cases does not exceed $2 \cdot 10^5$ .

For each test case, print a single integer — the sum of costs of all substrings of $s$ .

In the first test case, there is the only substring ")". Its cost is $1$ because we can insert '(' to the beginning of this substring and get a string "()", that is a balanced string.

In the second test case, the cost of each substring of length one is $1$ . The cost of a substring ")(" is $1$ because we can cyclically shift it to right and get a string "()". The cost of strings ")()" and "()(" is $1$ because its enough to insert one bracket to each of them. The cost of substring ")()(" is $1$ because we can cyclically shift it to right and get a string "()()". So there are $4 + 2 + 2 + 1 = 9$ substring of cost $1$ and $1$ substring of cost $0$ . So the sum of the costs is $9$ .

In the third test case,

• "(", the cost is $1$ ;
• "()", the cost is $0$ ;
• "())", the cost is $1$ ;
• ")", the cost is $1$ ;
• "))", the cost is $2$ ;
• ")", the cost is $1$ .

So the sum of the costs is $6$ .