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>
using namespace std;
char a[1000005];
int main() {
  int n;
  scanf("%d", &n);
  scanf("%s", a + 1);
  int numl = 0, numr = 0;
  int pos1, pos2;
  bool flag = 1;
  if (n % 2) flag = 0;
  for (int i = 1; i <= n; i++) {
    if (a[i] == '(')
      numl++;
    else {
      if (numl) {
        numl--;
      } else {
        numr++;
        if (numr == 1) {
          pos1 = i;
        }
        if (numr > 2) flag = 0;
      }
    }
  }
  int numl1 = 0, numr1 = 0;
  for (int i = n; i >= 1; i--) {
    if (a[i] == ')')
      numr1++;
    else {
      if (numr1)
        numr1--;
      else {
        numl1++;
        if (numl1 == 1) {
          pos2 = i;
        }
        if (numl1 > 2) flag = 0;
      }
    }
  }
  if (flag) {
    if (numr == 2 && !numl) {
      printf("%d\n", (pos1 - 1) / 2 + 1);
    } else if (numl1 == 2 && !numr1) {
      printf("%d\n", (n - pos2) / 2 + 1);
    } else {
      printf("0\n");
    }
  } else
    printf("0\n");
  return 0;
}
