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>
const long long MAX_N = 1e6 + 10;
const long long MOD = 1e9 + 7;
using namespace std;
int a[MAX_N];
int dp[MAX_N];
int main() {
  int n;
  string s;
  cin >> n >> s;
  if (n & 1) return cout << 0, 0;
  for (int i = 1; i <= n; ++i) {
    a[i] = a[i - 1];
    a[i] += (s[i - 1] == '(');
    a[i] -= (s[i - 1] == ')');
  }
  dp[n] = a[n];
  for (int i = n - 1; i > 0; --i) dp[i] = min(dp[i + 1], a[i]);
  long long ans = 0;
  if (s[n - 1] == '(') {
    s[n - 1] = ')';
    int now = 0;
    for (int i = 0; i < n; ++i) {
      now += (s[i] == '(');
      now -= (s[i] == ')');
      if (now < 0) return cout << 0, 0;
    }
    return cout << (!now), 0;
  }
  if (s[0] == ')') {
    s[0] = '(';
    int now = 0;
    for (int i = 0; i < n; ++i) {
      now += (s[i] == '(');
      now -= (s[i] == ')');
      if (now < 0) return cout << 0, 0;
    }
    return cout << (!now), 0;
  }
  for (int i = 2; i < n; ++i) {
    if (s[i - 1] == '(') {
      if (dp[i] >= 2 and a[n] == 2) ans++;
    } else {
      if (dp[i] >= -2 and a[n] == -2) ans++;
    }
    if (a[i] < 0) break;
  }
  cout << ans;
  return 0;
}
