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 N = 1e6 + 5;
bool A[N], poss[N];
int L[N], R[N], LL[N], RR[N];
int main() {
  ios_base ::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, l = 0, r = 0;
  cin >> n;
  for (int i = 0; i < n; i++) {
    char c;
    cin >> c;
    if (c == '(')
      A[i] = 1, l++;
    else
      r++;
    L[i] = l, R[i] = r;
  }
  l = 0, r = 0;
  for (int i = n - 1; i >= 0; i--) {
    if (A[i])
      l++;
    else
      r++;
    LL[i] = l, RR[i] = r;
  }
  if (abs(l - r) != 2) return cout << 0, 0;
  bool flag = 0;
  for (int i = 0; i < n; i++) {
    poss[i] = flag;
    if (R[i] > L[i]) flag = 1;
  }
  flag = 0;
  for (int i = n - 1; i >= 0; i--) {
    poss[i] |= flag;
    if (LL[i] > RR[i]) flag = 1;
  }
  int ans = 0;
  for (int i = 0; i < n; i++) {
    if (poss[i]) continue;
    if (A[i] && l > r) ans++;
    if (!A[i] && r > l) ans++;
  }
  cout << ans;
}
