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 long long inf = 1e17 + 10;
const int N = 1e6 + 10;
const long long mod = 998244353;
map<string, int> ml;
long long b[N], vis[N], po[N], num[N], t, n, m, x, y, k, a[N];
long long ex, ey, cnt, ans, sum, flag;
long long dist[N];
long long dp[N];
vector<int> v[N];
map<int, int> mp;
priority_queue<pair<int, int> > q;
priority_queue<pair<int, int> > tq;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  string s;
  cin >> s;
  for (int i = 0; s[i]; i++) {
    a[i] = cnt;
    if (s[i] == '(')
      cnt++;
    else
      cnt--;
    if (a[i - 1] < 0) a[i] = -1;
  }
  cnt = 0;
  for (int i = n - 1; i >= 0; i--) {
    b[i] = cnt;
    if (s[i] == ')')
      cnt++;
    else
      cnt--;
    if (b[i + 1] < 0) b[i] = -1;
  }
  for (int i = 0; s[i]; i++) {
    if (a[i] < 0 && b[i] < 0) return cout << 0 << endl, 0;
    num[i] = a[i] - b[i];
    if (s[i] == '(' && num[i] == 1 && a[i] >= 0 && b[i] >= 0) ans++;
    if (s[i] == ')' && num[i] == -1 && a[i] >= 0 && b[i] >= 0) ans++;
  }
  cout << ans << endl;
}
