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 aux, arr[1000100];
string pal;
stack<int> s;
pair<int, int> ans;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  ans.second = 1;
  cin >> pal;
  for (int i = 0; i < pal.size(); i++)
    if (pal[i] == '(') {
      s.push(i);
    } else if (!s.empty()) {
      arr[s.top()] = i;
      s.pop();
    }
  for (int i = 0; i < pal.size(); i++) {
    aux = 0;
    while (arr[i]) {
      aux += arr[i] - i + 1;
      i = arr[i] + 1;
    }
    if (aux > ans.first) {
      ans.first = aux, ans.second = 1;
    } else if (aux == ans.first && aux)
      ans.second++;
  }
  cout << ans.first << " " << ans.second;
}