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
from collections import deque
import math
input_ = lambda: sys.stdin.readline().strip("\r\n")
ii = lambda : int(input_())
il = lambda : list(map(int, input_().split()))
ilf = lambda : list(map(float, input_().split()))
ip = lambda : input_()
fi = lambda : float(input_())
li = lambda : list(input_())
pr = lambda x : print(x)
f = lambda : sys.stdout.flush()

n = ii()
s = li()
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])
#print(a)
if n%2 or len(a) == 0 :
    print(0)
elif 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)
else :
    print(0)
