Problem: This is yet another problem dealing with regular bracket sequences.

We should remind you that a bracket sequence is called regular, if by inserting «+» and «1» into it we can get a correct mathematical expression. For example, sequences «(())()», «()» and «(()(()))» are regular, while «)(», «(()» and «(()))(» are not. 

You are given a string of «(» and «)» characters. You are to find its longest substring that is a regular bracket sequence. You are to find the number of such substrings as well.

Input

The first line of the input file contains a non-empty string, consisting of «(» and «)» characters. Its length does not exceed 106.

Output

Print the length of the longest substring that is a regular bracket sequence, and the number of such substrings. If there are no such substrings, write the only line containing "0 1".

Examples

Input

)((())))(()())


Output

6 2


Input

))(


Output

0 1

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const long long mod = 1e9 + 7;
void solve() {
  string s;
  cin >> s;
  stack<pair<long long, long long> > st;
  map<long long, long long> mp;
  mp[0] = 1;
  for (long long i = 0; i < (long long)((s).size()); i++) {
    if (s[i] == '(') {
      st.push({0, -1});
    } else {
      if ((long long)((st).size()) == 0)
        continue;
      else if (st.top().second != -1) {
        st.pop();
      } else {
        long long f = st.top().first + 1;
        st.pop();
        if ((long long)((st).size()) == 0) {
          mp[f]++;
          st.push({f, 0});
        } else {
          st.top().first += f;
          mp[st.top().first]++;
        }
      }
    }
  }
  pair<long long, long long> ans;
  for (auto i : mp) {
    ans = i;
  }
  cout << ans.first * 2 << ' ' << ans.second << '\n';
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  while (t--) solve();
  return 0;
}