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 inf = 1e17;
const long long N = 1e5 + 100;
const long long bits = 21;
const long long mod = 1e9 + 7;
const long double pi = 3.14159265358979323846;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  std::cout << std::fixed << std::setprecision(20);
  ;
  string s;
  cin >> s;
  int n = s.size();
  int ans = 0;
  int cnt = 1;
  stack<pair<char, int>> st;
  for (int i = 0; i < n; i++) {
    if (st.empty()) {
      st.push({s[i], i});
    } else {
      pair<char, int> tp = st.top();
      if (tp.first == '(' && s[i] == ')') {
        st.pop();
        if (st.empty()) {
          ans = i + 1;
          cnt = 1;
        } else {
          tp = st.top();
          int sz = i - tp.second;
          if (sz == ans) {
            cnt++;
          } else if (sz > ans) {
            ans = sz;
            cnt = 1;
          }
        }
      } else {
        st.push({s[i], i});
      }
    }
  }
  cout << ans << " " << cnt << "\n";
}