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;
inline void boost() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
}
struct st {
  int open, close;
};
st d[int(4e6) + 1];
string a;
int n;
void built(int v, int l, int r) {
  if (l == r) {
    d[v].close = 0;
    d[v].open = 0;
    if (a[l - 1] == ')') {
      d[v].close = 1;
    } else {
      d[v].open = 1;
    }
  } else {
    int mid = (l + r) / 2;
    built(v * 2, l, mid);
    built(v * 2 + 1, mid + 1, r);
    d[v].open = d[v * 2].open - min(d[v * 2].open, d[v * 2 + 1].close) +
                d[v * 2 + 1].open;
    d[v].close = d[v * 2 + 1].close - min(d[v * 2].open, d[v * 2 + 1].close) +
                 d[v * 2].close;
  }
}
void update(int v, int l, int r, int pos) {
  if (l == r) {
    d[v].open = 0;
    d[v].close = 0;
    if (a[l - 1] == ')') {
      d[v].open = 1;
      a[l - 1] = '(';
    } else {
      d[v].close = 1;
      a[l - 1] = ')';
    }
  } else {
    int mid = (l + r) / 2;
    if (pos <= mid) {
      update(v * 2, l, mid, pos);
    } else {
      update(v * 2 + 1, mid + 1, r, pos);
    }
    d[v].open = d[v * 2].open - min(d[v * 2].open, d[v * 2 + 1].close) +
                d[v * 2 + 1].open;
    d[v].close = d[v * 2 + 1].close - min(d[v * 2].open, d[v * 2 + 1].close) +
                 d[v * 2].close;
  }
}
int main() {
  boost();
  cin >> n;
  cin >> a;
  int ans = 0;
  built(1, 1, n);
  for (int i = 0; i < n; i++) {
    update(1, 1, n, i + 1);
    if (d[1].open == 0 && d[1].close == 0) {
      ans++;
    }
    update(1, 1, n, i + 1);
  }
  cout << ans;
  return 0;
}
