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>
#pragma comment(linker, "/stack:200000000")
using namespace std;
const int N = 1e6 + 5, INF = 1e9 + 7;
int n, sz;
char a[N];
struct type {
  int open, close, res;
};
type t[N * 4];
type combine(type a, type b) {
  type ans;
  int t = min(a.open, b.close);
  ans.res = a.res + b.res + 2 * t;
  ans.open = a.open + b.open - t;
  ans.close = a.close + b.close - t;
  return ans;
}
inline void update(int v, type val) {
  t[v] = val;
  while (v > 0) {
    v /= 2;
    t[v] = combine(t[v * 2], t[v * 2 + 1]);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n;
  for (sz = 1; sz < n; sz *= 2)
    ;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
    if (a[i] == '(')
      update(i + sz - 1, {1, 0, 0});
    else
      update(i + sz - 1, {0, 1, 0});
  }
  int ans = 0;
  for (int i = 1; i <= n; ++i) {
    if (a[i] == '(')
      a[i] = ')', update(i + sz - 1, {0, 1, 0});
    else
      a[i] = '(', update(i + sz - 1, {1, 0, 0});
    if (t[1].res == n) ans++;
    if (a[i] == '(')
      a[i] = ')', update(i + sz - 1, {0, 1, 0});
    else
      a[i] = '(', update(i + sz - 1, {1, 0, 0});
  }
  cout << ans;
}
