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 + 2;
int n;
char s[N];
int prefixSum[N];
int minPref[N], minSuff[N];
int ToVal(char bracket) { return bracket == '(' ? 1 : -1; }
int main() {
  cin >> n >> s;
  for (int i = 0; i < n; ++i)
    prefixSum[i] = (i > 0 ? prefixSum[i - 1] : 0) + ToVal(s[i]);
  minPref[0] = prefixSum[0];
  for (int i = 1; i < n; ++i) minPref[i] = min(minPref[i - 1], prefixSum[i]);
  minSuff[n - 1] = prefixSum[n - 1];
  for (int i = n - 2; i >= 0; --i)
    minSuff[i] = min(minSuff[i + 1], prefixSum[i]);
  int ans = 0;
  for (int i = 0; i < n; ++i) {
    if (prefixSum[n - 1] - 2 * ToVal(s[i]) == 0 &&
        (i > 0 ? minPref[i - 1] >= 0 : true) &&
        minSuff[i] - 2 * ToVal(s[i]) >= 0)
      ++ans;
  }
  cout << ans;
  return 0;
}
