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;
int drz[2 * 1048576 + 5], suma, w, pref[1000005], ans, n, val, sumaa, val1,
    nr[1000005], nrr, prefl[1000005], rob[1000005];
int base = 1048575;
int t[1000005];
char c[1000005];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> c[i];
    if (c[i] == '(')
      t[i] = 1;
    else
      t[i] = -1;
    suma += t[i];
    pref[i] = pref[i - 1] + t[i];
    w = base + i;
    while (w >= 1) {
      drz[w] += t[i];
      w /= 2;
    }
  }
  for (int i = n; i >= 1; i--) {
    prefl[i] = prefl[i + 1] + t[i];
    if (prefl[i] <= 1 && nrr == 0) rob[i] = 1;
    if (prefl[i] > 0) nrr = 1;
  }
  for (int i = 1; i <= n; i++) {
    if (pref[i - 1] < 0) break;
    if (!rob[i]) continue;
    if (t[i] == 1)
      val = -1;
    else
      val = 1;
    if (pref[n] - pref[i] + val == -pref[i - 1] && suma + val * 2 == 0) ans++;
  }
  cout << ans;
}
