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;
string s;
int len, maxlen[1000005], maxs;
vector<int> kandidat;
int main() {
  cin >> s;
  len = ((int)(s).size());
  for (int i = (0); i < (len); ++i) {
    if (s[i] == '(') {
      maxlen[i] = -1;
      kandidat.push_back(i);
    } else {
      if (kandidat.empty()) {
        maxlen[i] = -1;
      } else {
        maxlen[i] = i - kandidat.back() + 1;
        if (kandidat.back() > 0 && maxlen[kandidat.back() - 1] > 0)
          maxlen[i] += maxlen[kandidat.back() - 1];
        kandidat.pop_back();
      }
    }
  }
  maxlen[len] = 0;
  maxs = *(max_element(maxlen, maxlen + len + 1));
  cout << maxs << " " << count(maxlen, maxlen + len + 1, maxs) << "\n";
  return 0;
}