#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define F first
#define S second
#define pii pair<int, int>
#define eb emplace_back
#define all(v) v.begin(), v.end()
#define rep(i, n) for (int i = 0; i < (n); ++i)
#define rep3(i, l, n) for (int i = l; i < (n); ++i)
#define sz(v) (int)v.size()
#define endl '\n'
const int inf = 1000000007;
const ll INF = 1e18;
int mod = 998244353;
// int mod = 1000000007;
#define abs(x) (x >= 0 ? x : -(x))
#define lb(v, x) (int)(lower_bound(all(v), x) - v.begin())
#define ub(v, x) (int)(upper_bound(all(v), x) - v.begin())
template<typename T1, typename T2> inline bool chmin(T1 &a, T2 b) { if (a > b) { a = b; return 1; } return 0; }
template<typename T1, typename T2> inline bool chmax(T1 &a, T2 b) { if (a < b) { a = b; return 1; } return 0; }
ll gcd(ll a, ll b) { if (b == 0) return a; return gcd(b, a % b); }
ll lcm(ll a, ll b) { return a / gcd(a, b) * b; }
template<typename T, typename U> T pow_(T a, U b) { return b ? pow_(a * a, b / 2) * (b % 2 ? a : 1) : 1; }
ll modpow(ll a, ll b, ll _mod) { return b ? modpow(a * a % _mod, b / 2, _mod) * (b % 2 ? a : 1) % _mod : 1; }
template<class T, class U> ostream& operator << (ostream& os, const pair<T, U>& p) { os << p.F << " " << p.S; return os; }
template<class T> ostream& operator << (ostream& os, const vector<T>& vec) { rep(i, sz(vec)) { if (i) os << " "; os << vec[i]; } return os; }
template<typename T> inline istream& operator >> (istream& is, vector<T>& v) { rep(j, sz(v)) is >> v[j]; return is; }
template<class T, class T2> inline void add(T &a, T2 b) { a += b; if (a >= mod) a -= mod; }


void solve();

int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cout << fixed << setprecision(10);
  int T;
  T = 1;
  while (T--) solve();
}

// https://kmjp.hatenablog.jp/entry/2018/05/13/0900
// https://atcoder.jp/contests/abc167/tasks/abc167_f
void solve() {
  int n;
  cin >> n;
  string s;
  cin >> s;

  vector<pii> l, r;
  {
    int now = 0, mi = 0;
    rep(i, n) {
      if (s[i] == ')') { now--; chmin(mi, now); }
      else now++;
      if (now >= 0) l.eb(mi, now);
      else r.eb(mi, now);
    }
  }

  sort(l.rbegin(), l.rend());
  sort(all(r));

  {
    int mi = 0, now = 0;
    rep(i, sz(l)) {
      // cout << l[i] << endl;
      chmin(mi, now + l[i].F);
      now += l[i].S;
    }
    rep(i, sz(r)) {
      // cout << r[i] << endl;
      chmin(mi, now + r[i].F);
      now += r[i].S;
    }
    // cout << mi << " " << now << endl;

    // 山すべてを除いた谷 1 つ )))...(( が余り
    // ))...) は mi
    // 谷 1 つだから mi ≤ 0
    // ) の個数は -mi
    // ( の個数は now-mi
    // それらを引く
    cout << 1LL * n * (n + 1) / 2 + mi - (now - mi) << endl;
  }
}