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> st;
  vector<int> d(int((s).size()), -1), c(int((s).size()), -1);
  for (int i = 0; i < int((s).size()); i++) {
    if (s[i] == '(') {
      st.push(i);
    } else {
      if (int((st).size())) {
        d[i] = st.top();
        st.pop();
        if (d[i] - 1 >= 0 and c[d[i] - 1] != -1) {
          c[i] = c[d[i] - 1];
        } else {
          c[i] = d[i];
        }
      }
    }
  }
  int count = 1, mx = 0;
  for (int i = 0; i < int((s).size()); i++) {
    if (c[i] != -1) {
      int len = i - c[i] + 1;
      if (mx == len) {
        count++;
      }
      if (mx < len) {
        mx = len;
        count = 1;
      }
    }
  }
  cout << mx << " " << count << endl;
  return 0;
}