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()
r = [0]*(n+1)
opening = 0
rem1 = [0]*(n+1)
r1 = 0
for i in range(n):
	if(s[i] == '('):
		opening += 1
	else:
		if(opening > 0):
			r1 += 2
			opening -= 1
	r[i+1] = opening
	rem1[i+1] = r1
closing = 0
l = [0]*(n+1)
rem2 = [0]*(n+1)
r1 = 0
for i in range(n-1, 0, -1):
	if(s[i] == ')'):
		closing += 1
	else:
		if(closing > 0):
			r1 += 2
			closing -= 1
	l[i-1] = closing
	rem2[i-1] = r1
if(abs(s.count(')')-s.count('(')) != 2):
	print(0)
	exit()
# print(r)
# print(l)
# print(rem1)
# print(rem2)

ans = 0
for i in range(n):
	if(abs(r[i]-l[i]) == 1 and r[i]+l[i]+rem2[i]+rem1[i] == n-1):
		ans += 1
print(ans)