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 main() {
  string s;
  cin >> s;
  stack<int> l;
  vector<int> x(s.size());
  for (int i = 0; i < s.size(); ++i) {
    if (l.empty() && s[i] == ')') continue;
    if (s[i] == '(') l.push(i);
    if (s[i] == ')') {
      x[i] = 1;
      x[l.top()] = 1;
      l.pop();
    }
  }
  map<int, int> m;
  int t = 0;
  for (int i = 0; i < x.size(); ++i) {
    if (x[i] == 0) {
      m[-1 * t]++;
      t = 0;
    } else
      t++;
  }
  m[-1 * t]++;
  for (map<int, int>::iterator it = m.begin(); it != m.end(); ++it) {
    if (it->first != 0)
      cout << -1 * it->first << ' ' << it->second << endl;
    else
      cout << "0 1" << endl;
    break;
  }
  return 0;
}