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 good[1000010];
int main() {
  string s;
  cin >> s;
  int len = s.size();
  fill_n(good, len + 1, 1);
  int i = 0;
  int e = 0;
  stack<int> st;
  while (i < s.size()) {
    if (s[i] == '(') {
      st.push(i);
    } else {
      if (!st.empty())
        st.pop();
      else
        good[i] = false;
    }
    i++;
  }
  while (!st.empty()) {
    good[st.top()] = false;
    st.pop();
  }
  int t = 0, max = 0, k = 1;
  good[s.size()] = false;
  for (int i = 0; i <= s.size(); i++) {
    if (good[i] == 1) {
      t++;
    } else {
      if (t > max) {
        max = t;
        k = 0;
      }
      if ((t > 0) and (t == max)) k++;
      t = 0;
    }
  }
  cout << max << " " << k;
}