You are given a bracket sequence s consisting of n opening '(' and closing ')' brackets.

A regular bracket sequence is a bracket sequence that can be transformed into a correct arithmetic expression by inserting characters '1' and '+' between the original characters of the sequence. For example, bracket sequences "()()", "(())" are regular (the resulting expressions are: "(1)+(1)", "((1+1)+1)"), and ")(" and "(" are not.

You can change the type of some bracket s_i. It means that if s_i =  ')' then you can change it to '(' and vice versa.

Your task is to calculate the number of positions i such that if you change the type of the i-th bracket, then the resulting bracket sequence becomes regular.

Input

The first line of the input contains one integer n (1 ≤ n ≤ 10^6) — the length of the bracket sequence.

The second line of the input contains the string s consisting of n opening '(' and closing ')' brackets.

Output

Print one integer — the number of positions i such that if you change the type of the i-th bracket, then the resulting bracket sequence becomes regular.

Examples

Input


6
(((())


Output


3


Input


6
()()()


Output


0


Input


1
)


Output


0


Input


8
)))(((((


Output


0
n = int(input())
S = input()

def solve(s, counts):
    openCount = 0
    closeCount = 0
    posCount = 0
    if n % 2 == 1:
        return posCount
    for i in range(n):
        if s[i] == '(':
            openCount += 1
        else:
            closeCount += 1
        if closeCount - openCount > 2:
            return posCount
        else:
            counts.append([openCount, closeCount])
    if abs(openCount-closeCount) == 2:
        if openCount > closeCount:
            s = s[::-1]
            s = ['(' if c == ')' else ')' for c in s]
            counts = []
            openCount = 0
            closeCount = 0
            for i in range(n):
                if s[i] == '(':
                    openCount += 1
                else:
                    closeCount += 1
                if closeCount - openCount > 2:
                    return posCount
                else:
                    counts.append([openCount, closeCount])
        for i in range(n):
            if counts[i][1] - counts[i][0] < 2 and s[i] == ')':
                posCount += 1
            if counts[i][1] > counts[i][0]:
                break
        return posCount
    else:
        return posCount

print(solve(S,[]))