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;
void printVector(vector<int> &a) {
  for (vector<int>::iterator it = a.begin(); it != a.end(); it++)
    cout << *it << " ";
  cout << "\n";
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  string s;
  cin >> s;
  vector<int> a, b(s.size() + 1, 0);
  int i, maxi = 0, count = 1, k;
  for (i = 0; i < s.size(); i++) {
    if (s[i] == ')' && a.size() && s[a.back()] == '(') {
      k = (i - a.back() + 1);
      b[i + 1] = (i - a.back() + 1) + b[a.back()];
      if (maxi < b[i + 1]) {
        maxi = b[i + 1];
        count = 1;
      } else if (maxi == b[i + 1])
        count++;
      a.pop_back();
    } else
      a.push_back(i);
  }
  cout << maxi << " " << count;
  return 0;
}