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, ls = int(input()), [1 if x == "(" else -1 for x in input()]
op = sum(ls)
if n & 1 or op > 2 or op < -2 or op == 0:
	print(0)
	raise SystemExit()
runningsum = 0
cd = False
poss = 0
if op > 0:
	for i in range(n):
		runningsum += ls[i]
		if runningsum > 1 and ls[i] == 1:
			poss+=1
		if runningsum < 2: poss = 0
		if runningsum < 0:
			print(0)
			raise SystemExit()
else:
	for i in range(n-1, -1, -1):
		runningsum-=ls[i]
		if runningsum > 1 and ls[i] == -1:
			poss+=1
		if runningsum < 2: poss = 0
		if runningsum < 0:
			print(0)
			raise SystemExit()
print(poss)