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 main():
    n = int(input())
    s = input()
    if n == 1:
        print(0)
        return
    dpleft = []
    dpright = []
    for i in range(n):
        dpleft.append([0,0])
        dpright.append([0,0])


    left = 0
    right = 0

    for i in range(n):
        if s[i] == '(':
            left += 1
        else:
            if left > 0:
                left -= 1
            else:
                right += 1

        dpleft[i][0] = left
        dpleft[i][1] = right

    left = 0
    right = 0

    for i in range(n-1,-1,-1):
        if s[i] == ')':
            right += 1
        else:
            if right > 0:
                right -= 1
            else:
                left += 1

        dpright[i][0] = left
        dpright[i][1] = right

    ans = 0
    #print(dpleft,dpright)

    for i in range(n):
        if i == 0:
            if dpright[i+1][0] == 0:
                left = 0
                right = dpright[i+1][1]
                if s[i] == ')':
                    left += 1
                else:
                    right += 1
                if left == right:
                    ans += 1
        elif i == n-1:
            if dpleft[i-1][1] == 0:
                left = dpleft[i-1][0]
                right = 0
                if s[i] == ')':
                    left += 1
                else:
                    right += 1
                if left == right:
                    ans += 1
        else:
            if dpleft[i-1][1] == 0 and dpright[i+1][0] == 0:
                left = dpleft[i-1][0]
                right = dpright[i+1][1]
                if s[i] == ')':
                    left += 1
                else:
                    right += 1
                if left == right:
                    ans += 1

    print(ans)


main()
