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()

diff, cura, curb = [], 0, 0
for item in s:
    if item == '(':
        cura += 1
    else:
        curb += 1
    diff.append(cura - curb)

if diff[-1] == 2:
    if min(diff) < 0:
        print(0)
    else:
        ans, i = 0, n - 1
        while i >= 0 and diff[i] >= 2:
            if s[i] == '(':
                ans += 1
            i -= 1
        print(ans)
elif diff[-1] == -2:
    if min(diff) < -2:
        print(0)
    else:
        ans = 0
        for i in range(0, n-1):
            if s[i] == ')' and diff[i] >= 0:
                ans += 1
            elif diff[i] < 0:
                if s[i] == ')':
                    ans += 1
                break
        print(ans)
else:
    print(0)