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;
const int oo = 0x3f3f3f3f;
const long long ooo = 9223372036854775807ll;
const int _cnt = 1000 * 1000 + 7;
const int _p = 1000 * 1000 * 1000 + 7;
const int N = 1000005;
const double PI = acos(-1.0);
const double eps = 1e-9;
int o(int x) { return x % _p; }
int gcd(int a, int b) { return b ? gcd(b, a % b) : a; }
int lcm(int a, int b) { return a / gcd(a, b) * b; }
void file_put() {
  string s = "/home/jslijin/jslijin/code/";
  freopen((s + "code.in").c_str(), "r", stdin);
  freopen((s + "code.out").c_str(), "w", stdout);
}
int n, a[N], p[N], s[N], ans = 0;
char st[N];
int main() {
  scanf("%d%s", &n, st + 1), p[n + 1] = oo;
  for (int i = (1); i <= (n); ++i)
    a[i] = (st[i] == '(') ? 1 : -1, s[i] = s[i - 1] + a[i];
  for (int i = (n); i >= (1); --i)
    p[i] = ((p[i + 1]) < (s[i]) ? (p[i + 1]) : (s[i]));
  if (abs(s[n]) != 2) return 0 * printf("0\n");
  for (int i = (1); i <= (n); ++i) {
    if (s[i - 1] < 0) break;
    if (a[i] * s[n] < 0 || p[i] < 2 * a[i]) continue;
    ans++;
  }
  printf("%d\n", ans);
  return 0;
}
