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;
bool matched[1000005];
int main() {
  string s;
  getline(cin, s);
  stack<pair<int, char> > myStack;
  for (int i = 0; i < s.size(); i++) {
    matched[i] = false;
    if (s[i] == ')' && !myStack.empty()) {
      pair<int, char> temp = myStack.top();
      myStack.pop();
      matched[i] = true;
      matched[temp.first] = true;
    } else if (s[i] == '(')
      myStack.push(make_pair(i, '('));
  }
  int max_length = 0, max_count = 0, current = 0;
  for (int i = 0; i < s.size(); i++) {
    if (matched[i])
      current++;
    else {
      if (current > max_length) {
        max_length = current;
        max_count = 1;
      } else if (current == max_length) {
        max_count++;
      }
      current = 0;
    }
  }
  if (current > max_length) {
    max_length = current;
    max_count = 1;
  } else if (current == max_length) {
    max_count++;
  }
  if (max_length == 0) {
    cout << "0 1" << endl;
  } else {
    cout << max_length << " " << max_count << endl;
  }
}