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;
  int freq[s.size() + 3], e[s.size() + 3], ss[s.size() + 3];
  for (int i = 0; i <= s.size(); i++) {
    freq[i] = 0;
    e[i] = -1;
  }
  for (int i = 1; i <= s.size(); i++) {
    if (s[i - 1] == '(') {
      st.push(i);
    }
    if (s[i - 1] == ')' and !st.empty()) {
      int x = st.top();
      st.pop();
      ss[i] = x;
      e[i] = x;
      if (e[e[i] - 1] != -1) e[i] = e[e[i] - 1];
      freq[i - e[i] + 1]++;
    }
  }
  int index = -1;
  for (int i = s.size(); i >= 0; i--) {
    if (freq[i] > 0) {
      cout << (i) << " " << freq[i] << endl;
      return 0;
    }
  }
  cout << 0 << " " << 1 << endl;
}