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(0);
  cin.tie(0);
  cout.tie(0);
  string s;
  cin >> s;
  stack<int> st;
  int size[1000005];
  vector<pair<int, int>> r;
  for (int i = 0; i < 1000005; i++) size[i] = 0;
  for (int i = 0; i < s.size(); i++) {
    if (s[i] == '(') {
      st.push(i);
    } else {
      if (st.size()) {
        r.push_back({st.top(), i});
        size[i] = i - st.top() + 1;
        st.pop();
      }
    }
  }
  int max = 0, cnt = 1;
  for (int i = 0; i < r.size(); i++) {
    if (r[i].first - 1 > 0)
      if (size[r[i].first - 1]) {
        size[r[i].second] += size[r[i].first - 1];
      }
    if (size[r[i].second] == max) cnt++;
    if (size[r[i].second] > max) {
      max = size[r[i].second];
      cnt = 1;
    }
  }
  cout << max << " " << cnt;
}