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 pasangan[1000005];
string s;
stack<int> brac;
map<int, int> sub;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> s;
  memset(pasangan, -1, sizeof(pasangan));
  for (int i = 0; i < s.size(); i++) {
    if (s[i] == '(') {
      brac.push(i);
    } else {
      if (brac.empty()) continue;
      pasangan[i] = brac.top();
      brac.pop();
      if (pasangan[i] && pasangan[pasangan[i] - 1] != -1) {
        pasangan[i] = pasangan[pasangan[i] - 1];
      }
      sub[i - pasangan[i] + 1]++;
    }
  }
  sub[0]++;
  cout << sub.rbegin()->first << ' ' << sub.rbegin()->second << '\n';
  return 0;
}