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
def isregular(s):
    ls=0
    rs=0
    for char in s:
        if char=="(":
            ls+=1
        else:
            rs+=1
        if rs>ls:
            return False
    if ls==rs:
        return True
    return False
n=int(input())
s=input()
if n%2==1:
    print(0)
else:
    a=s.count("(")
    b=n-a
    diff=[0]
    for i in range(n):
        if s[i]=="(":
            diff.append(diff[-1]+1)
        else:
            diff.append(diff[-1]-1)
    if a==b+2:
        if -1 in diff:
            print(0)
        else:
            backdiff=diff[::-1]
            last=backdiff.index(1)
            print(s[-last:].count("("))
    elif b==a+2:
        if -3 in diff:
            print(0)
        else:
            first=diff.index(-1)
            print(s[:first].count(")"))
    else:
        print(0)