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
import sys
import bisect
from random import randint
input=lambda: sys.stdin.readline().strip("\r\n")
from math import gcd
sa=lambda :input()
sb=lambda:int(input())
sc=lambda:input().split()
sd=lambda:list(map(int,input().split()))
se=lambda:float(input())
sf=lambda:list(input())
flsh=lambda: sys.stdout.flush()
mod=10**9+7
def hnbhai():
    n=sb()
    s=sa()
    a=[]
    for i in range(n):
        if s[i]==")" and len(a)>0 and a[-1][0]=="(":
            a.pop()
        else:
            a.append([s[i],i+1])
    if n%2 or len(a)==0:
        print(0)
        return
    if len(a)==2 and a[0][0]==a[1][0]:
        if a[0][0]=="(":
            print((n-a[1][1])//2+1)
        else:
            print(a[0][1]//2+1)
        return
    print(0)
for _ in range(1):
    hnbhai()
