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;
  int size = s.length();
  int a[1000002] = {0};
  std::stack<int> st;
  for (int i = 0; i < size; i++) {
    if (s[i] == '(') {
      st.push(i);
    } else {
      if (!st.empty()) {
        int tp = st.top();
        st.pop();
        a[tp] = 1;
        a[i] = 1;
      }
    }
  }
  int temp = 0, mx = 0, count = 0;
  for (int i = 0; i < size; i++) {
    if (a[i] == 1) {
      a[i] += (i - 1 >= 0 ? a[i - 1] : 0);
      mx = max(mx, a[i]);
    }
  }
  for (int i = 0; i < size; i++) {
    if (a[i] == mx && mx != 0) {
      count++;
    }
  }
  if (mx != 0)
    cout << mx << " " << count;
  else
    cout << "0"
         << " "
         << "1";
  return 0;
}