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;
const long long int mxsize = 3e5 + 5;
const long long int linf = 1e9;
const long double inf = 10000000000.0;
const long long int mod = 1e9 + 7;
int main() {
  string s;
  cin >> s;
  stack<int> st;
  st.push(-1);
  unordered_map<int, int> mp;
  int ans = 0;
  int n = s.length();
  for (int i = 0; i < n; i++) {
    if (s[i] == '(')
      st.push(i);
    else {
      st.pop();
      if (!st.empty()) {
        ans = i - st.top();
        mp[ans]++;
      } else
        st.push(i);
    }
  }
  int max_val = 0;
  for (auto x : mp) {
    max_val = max(max_val, x.first);
  }
  if (max_val == 0)
    cout << 0 << " " << 1 << endl;
  else
    cout << max_val << " " << mp[max_val] << endl;
}