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 last = -1;
int a[1000000];
stack<pair<char, int>> st;
void khali() {
  for (int i = 0; i < st.size(); i++) st.pop();
}
int main() {
  string s;
  getline(cin, s);
  for (int i = 0; i < s.size(); i++) {
    if (st.size() == 0 && s[i] == '(') {
      st.push({'(', i});
    } else if (st.size() == 0 && s[i] == ')') {
      last = i;
    } else if (st.size() != 0 && s[i] == '(') {
      st.push({'(', i});
    } else if (st.size() != 0 && s[i] == ')') {
      if (st.top().first == '(') {
        st.pop();
        if (st.size() != 0) {
          a[st.top().second + 1] = i - st.top().second;
        } else {
          a[last + 1] = i - last;
        }
      } else {
        khali();
        last = i;
      }
    }
  }
  int cnt = 0;
  int Max = 0;
  for (int i = 0; i < 1000000; i++) {
    if (a[i] > Max) {
      Max = a[i];
      cnt = 1;
    } else if (a[i] == Max) {
      cnt++;
    }
  }
  if (Max == 0) {
    cout << 0 << " " << 1;
    return 0;
  }
  cout << Max << " " << cnt;
}