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() )
l = list(input() )
cptMoins = 0
cptPlus = 0
cpt = 0
dead = False
plusEnd = False
for k in range (n):
	if(l[k]=='('):
		cpt+=1
		cptMoins+=1
	else:
		cpt-=1
		if(not plusEnd):
			cptPlus+=1
	if( cpt < -2): # cas de fin
		dead = True
		break
	if ( cpt < 0): #on doit utiliser un joker avant
		if(plusEnd or cptPlus==0):#on en a pas ou déjà utilisé
			dead = True
			break
		else:
			plusEnd = True
			#on considère qu'on l'a fait
			cpt +=2

	if cpt<2:
		cptMoins =0
if(dead):
	print(0)
else:
	if( plusEnd and cpt ==0):
		print(cptPlus)
	elif ( not plusEnd and cpt ==2):
		print(cptMoins)
	else:
		print(0)
