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())
xs = list(map(lambda x: 1 if x == '(' else -1, list(input())))

# total sum 0
# never under 0

head = [0 for _ in range(0, n + 1)]
tail = [0 for _ in range(0, n + 1)]

head[0] = xs[0]

for i in range(1, n):
    head[i] = head[i - 1] + xs[i]

tail[n] = 0
for i in range(n - 1 , 0, -1):
    tail[i] = tail[i + 1] + xs[i]

a = 0
for x in head[:-1]:
    a += 1
    if x < 0:
        break

b = 0
for x in reversed(tail[:-1]):
    b += 1
    if x > 0:
        break

ans = 0
for i in range(len(xs) - b, a):
    diff = 2 if xs[i] < 0 else -2
    if head[i] + diff >= 0 and head[i] + diff + tail[i + 1] == 0:
        ans += 1

print(ans)
