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 warning(disable : 6031)
#pragma warning(disable : 4101)
using namespace std;
const int INF = 1e9;
const long long LINF = 1e18;
const double PI = acos(-1);
const int MAXN = 1000005;
int n, tot, Z[MAXN], DP[MAXN];
char S[MAXN];
int main() {
  scanf("%d", &n);
  scanf(" %s", S);
  Z[0] = S[0] == '(' ? 1 : -1;
  for (int i = 1; i < n; i++) {
    Z[i] = Z[i - 1] + (S[i] == '(' ? 1 : -1);
  }
  DP[n - 1] = Z[n - 1];
  for (int i = n - 2; i >= 0; i--) {
    DP[i] = min(DP[i + 1], Z[i]);
  }
  for (int i = 0; i < n; i++) {
    if (Z[n - 1] == 2 && S[i] == '(' && DP[i] >= 2) {
      tot++;
    } else if (Z[n - 1] == -2 && S[i] == ')' && DP[i] >= -2) {
      tot++;
    }
    if (Z[i] < 0) {
      break;
    }
  }
  printf("%d\n", tot);
  return 0;
}
