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 const N = 1e6 + 5;
int main() {
  string a;
  int length = 0, subStrings = 1, mx = -1;
  cin >> a;
  stack<int> brackets;
  brackets.push(-1);
  int t = a.size(), Freq[N] = {};
  for (int i = 0; i < t; i++) {
    if (a[i] == '(')
      brackets.push(i);
    else {
      brackets.pop();
      if (brackets.size()) {
        length = i - brackets.top();
        Freq[length]++;
      } else
        brackets.push(i);
    }
  }
  for (int i = N - 1; i >= 0; i--) {
    if (Freq[i]) {
      length = i;
      subStrings = Freq[i];
      break;
    }
  }
  cout << length << " " << subStrings << endl;
  return 0;
}