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 ii():
    return int(input())
def mi():
    return map(int, input().split())
def li():
    return list(mi())

n = ii()
s = input().strip()
a = [0] * (n + 1)
m = [0] * (n + 1)
for i in range(n):
    a[i] = a[i - 1] + (1 if s[i] == '(' else -1)
    m[i] = min(m[i - 1], a[i])

ans = 0
mm = a[n - 1]
for j in range(n - 1, -1, -1):
    mm = min(mm, a[j])
    if s[j] == '(':
        if a[n - 1] == 2 and mm == 2 and m[j - 1] >= 0:
            ans += 1
    else:
        if a[n - 1] == -2 and mm == -2 and m[j - 1] >= 0:
            ans += 1

print(ans)
