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1374C. Move Brackets ​

greedy, string, 1000, https://codeforces.com/problemset/problem/1374/C

You are given a bracket sequence 𝑠 of length 𝑛, where 𝑛 is even (divisible by two). The string 𝑠 consists of 𝑛2 opening brackets '(' and 𝑛2closing brackets ')'.

In one move, you can choose exactly one bracket and move it to the beginning of the string or to the end of the string (i.e. you choose some index 𝑖, remove the 𝑖-th character of 𝑠 and insert it before or after all remaining characters of 𝑠).

Your task is to find the minimum number of moves required to obtain regular bracket sequence from 𝑠. It can be proved that the answer always exists under the given constraints.

Recall what the regular bracket sequence is:

  • "()" is regular bracket sequence;
  • if 𝑠 is regular bracket sequence then "(" + 𝑠 + ")" is regular bracket sequence;
  • if 𝑠 and 𝑡 are regular bracket sequences then 𝑠 + 𝑡 is regular bracket sequence.

For example, "()()", "(())()", "(())" and "()" are regular bracket sequences, but ")(", "()(" and ")))" are not.

You have to answer 𝑡 independent test cases.

Input

The first line of the input contains one integer 𝑡 (1≤𝑡≤2000) — the number of test cases. Then 𝑡 test cases follow.

The first line of the test case contains one integer 𝑛 (2≤𝑛≤50) — the length of 𝑠. It is guaranteed that 𝑛 is even. The second line of the test case containg the string 𝑠 consisting of 𝑛2 opening and 𝑛2 closing brackets.

Output

For each test case, print the answer — the minimum number of moves required to obtain regular bracket sequence from 𝑠. It can be proved that the answer always exists under the given constraints.

Example

input

4
2
)(
4
()()
8
())()()(
10
)))((((())

output

1
0
1
3

Note

In the first test case of the example, it is sufficient to move the first bracket to the end of the string.

In the third test case of the example, it is sufficient to move the last bracket to the beginning of the string.

In the fourth test case of the example, we can choose last three openning brackets, move them to the beginning of the string and obtain "((()))(())".

思路：遍历字符串时维护一个平衡值 bal（遇到 ( 加 1，遇到 ) 减 1）。

  • 如果 bal 从未小于 0，说明序列是合法的。
  • 如果 bal 出现过负数，最小值 min_bal 就表示右括号超前多少个。 答案就是 -min_bal（需要把多少个右括号挪到后面）。

代码

python
t = int(input())
for _ in range(t):
    n = int(input())
    s = input().strip()
    bal = 0
    min_bal = 0
    for ch in s:
        bal += 1 if ch == '(' else -1
        min_bal = min(min_bal, bal)
    print(-min_bal)