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().strip()
a=[0]*(n+5)
a2=[0]*(n+5)
mn=[0]*(n+5)
mn2=[0]*(n+5)
for i in range(1,n+1,1):
    if s[i-1]=='(':
        a[i]=a[i-1]+1
    else:
        a[i]=a[i-1]-1
    mn[i]=min(mn[i-1],a[i])
    
for i in range(n,0,-1):
    if s[i-1]==')':
        a2[i]=a2[i+1]+1
    else:
        a2[i]=a2[i+1]-1
    mn2[i]=min(mn2[i+1],a2[i])
ans=0
if a[n]!=2 and a[n]!=-2:
    print(0);
    exit(0)
for i in range(1,n+1,1):
    if s[i-1]=='(' and a[n]==2:
        if mn[i-1]>=0 and mn2[i+1]>=0 and a[i-1]>=1:
            ans=ans+1
    elif s[i-1]==')' and a[n]==-2:
        if mn[i-1]>=0 and mn2[i+1]>=0 and a2[i+1]>=1:
            ans=ans+1
        
print(ans)
    
