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()

pref=[0]*(n+1)
suf=[0]*(n+1)
okp=[0]*(n+1)
oks=[0]*(n+1)
okp[0]=1
oks[n]=1

def x(i):
    if s[i]=='(': return 1
    else: return -1
    
def y(i):
    if s[n-i-1]==')': return 1
    else: return -1

for i in range(n):
    pref[i+1] = pref[i] + x(i)
    okp[i+1] = okp[i] & (pref[i+1] >= 0)
    suf[n-i-1] = suf[n-i] + y(i)
    oks[n-i-1] = oks[n-i] & (suf[n-i-1] >= 0)

ans=0

for i in range(n):
    if (okp[i] and oks[i+1]) == 0 : continue
    if s[i]=='(':
        if pref[i]>0 and (pref[i]-1-suf[i+1] == 0):
            ans+=1
    else:
        if (pref[i]+1-suf[i+1]) == 0:
            ans+=1
        
print(ans)
        

