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
from sys import stdin

rstr = lambda: stdin.readline().strip()
n, s, cum, ans = int(input()), rstr(), [0], 0

for i in s:
    if i == '(':
        cum.append(cum[-1] + 1)
    else:
        cum.append(cum[-1] - 1)

mi = [cum[-1]]
for i in range(n - 1, -1, -1):
    mi.append(min(mi[-1], cum[i]))

mi.reverse()
if cum[-1] == 0 or abs(cum[-1]) != 2 or mi[0] < -2:
    print(0)
else:
    for i in range(n):
        if s[i] == ')':
            if cum[-1] + 2 == 0:
                ans += 1
        else:
            if cum[-1] - 2 == 0 and i and mi[i + 1] - 2 > -1:
                ans += 1

        if cum[i + 1] < 0:
            break

    print(ans)
