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()

if n % 2 == 1:
    print(0)
else:
    a, b, diff = [], [], []
    cura, curb = 0, 0
    for item in s:
        if item == '(':
            cura += 1
        else:
            curb += 1
        a.append(cura)
        b.append(curb)
        cur_diff = cura - curb
        diff.append(cur_diff)
    if a[-1] - b[-1] == 2:
        if min(diff) < 0:
            print(0)
        else:
            ans = 0
            for i in range(n-1, -1, -1):
                if diff[i] < 2:
                    break
                if s[i] == '(' and diff[i] >= 2:
                    ans += 1
            print(ans)
    elif b[-1] - a[-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)