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
def Rd(f = int):
  return [f(x) for x in input().split()]

def PrintFloats(arr):
  return ' '.join('{:.10f}'.format(x) for x in arr)

###

n = int(input())
s = input()

op = s.count('(')
cl = s.count(')')

def Solve(s):
  bal = 0
  end = n-1  
  for bwd in range(n-1, -1, -1):
    if s[bwd] == '(': bal += 1
    else: bal -= 1
    end = bwd
    if bal > 0: break

  bal = 0
  ans = 0
  for fwd in range(n):
    if s[fwd] == '(':
      if bal > 0 and fwd >= end: ans += 1
      bal += 1
    else: bal -= 1
    if bal < 0: break    
  return ans

if op == cl + 2:
  print(Solve(s))
elif cl == op + 2:
  s2 = s.replace('(', 'x').replace(')', '(').replace('x', ')')
  print(Solve(s2[::-1]))
else:
  print(0)
