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 maxn = 1000010;
int n, ans = 0;
string s;
int suf[maxn], pref[maxn];
bool pcan[maxn], scan[maxn], ok;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  cin >> s;
  ok = true;
  pcan[0] = 1;
  for (int i = 1; i <= n; i++) {
    pref[i] = (s[i - 1] == '(' ? 1 : -1);
    pref[i] += pref[i - 1];
    if (pref[i] < 0) ok = false;
    pcan[i] = ok;
  }
  ok = true;
  scan[n + 1] = 1;
  for (int i = n; i >= 1; i--) {
    suf[i] = (s[i - 1] == '(' ? 1 : -1);
    suf[i] += suf[i + 1];
    if (suf[i] > 0) ok = false;
    scan[i] = ok;
  }
  for (int i = 1; i <= n; i++) {
    if (!pcan[i - 1] || !scan[i + 1]) continue;
    if ((pref[i - 1] + suf[i + 1] + (s[i - 1] == '(' ? -1 : 1)) == 0) ans++;
  }
  cout << ans;
  return 0;
}
