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() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  string s;
  cin >> s;
  stack<string> st;
  int ans = 0;
  int c = 0;
  for (int i = 0; i < s.size(); i++) {
    if (s[i] == '(') {
      st.push("(");
    } else {
      int x = 0;
      while (!st.empty() && isdigit(st.top()[0])) {
        string t = st.top();
        int n = stoi(t);
        x += n;
        st.pop();
      }
      if (st.empty()) {
        continue;
      }
      st.pop();
      x += 2;
      while (!st.empty() && isdigit(st.top()[0])) {
        string t = st.top();
        int n = stoi(t);
        x += n;
        st.pop();
      }
      st.push(to_string(x));
      if (ans < x) {
        ans = x;
        c = 1;
      } else if (ans == x) {
        c++;
      }
    }
  }
  if (ans == 0) {
    cout << ans << " " << 1;
    return 0;
  }
  cout << ans << " " << c << "\n";
  return 0;
}