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
#include <bits/stdc++.h>
char s[1000010];
int c[1000010];
int n;
int min(int a, int b) { return a < b ? a : b; }
int check() {
  int i, low, ret;
  scanf("%d\n%s", &n, s);
  if (n & 1) return 0;
  c[0] = 0;
  low = 0;
  for (i = 0; i < n; i++) {
    if (s[i] == '(')
      c[i + 1] = c[i] + 1;
    else
      c[i + 1] = c[i] - 1;
    low = min(low, c[i + 1]);
  }
  if (c[n] == 2) {
    if (low < 0) return 0;
    ret = 0;
    for (i = n; i > 0; i--) {
      if (c[i] < 2) break;
      if (s[i - 1] == '(') ret++;
    }
    return ret;
  } else if (c[n] == -2) {
    if (low < -2) return 0;
    ret = 0;
    for (i = 0; i < n; i++) {
      if (c[i] < 0) break;
      if (s[i] == ')') ret++;
    }
    return ret;
  } else {
    return 0;
  }
}
int main() {
  int r;
  r = check();
  printf("%d", r);
}
