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;
int main() {
  string s;
  cin >> s;
  stack<long long> st;
  unordered_map<long long, long long> mp;
  vector<long long> l(s.length(), 0);
  for (long long i = 0; i < s.length(); i++) {
    if (s[i] == ')') {
      if (!st.empty()) {
        int len = i - st.top() + 1;
        if (st.top() != 0)
          l[i] = len + l[i - len];
        else
          l[i] = len;
        mp[l[i]]++;
        st.pop();
      }
    } else
      st.push(i);
  }
  long long n = s.length();
  mp[0] = 1;
  cout << *max_element(l.begin(), l.end()) << " "
       << mp[*max_element(l.begin(), l.end())];
  return 0;
}