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::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  int T = 1;
  for (int test_case = 1; test_case <= T; ++test_case) {
    string s;
    cin >> s;
    stack<int> open;
    int n = (int)s.size();
    int brackets[n + 1];
    for (int i = 0; i < n; ++i) brackets[i] = INT_MAX;
    for (int i = 0; i < n; ++i) {
      if (s[i] == ')' and !open.empty()) {
        brackets[i] = open.top();
        open.pop();
      } else {
        if (s[i] == ')')
          brackets[i] = INT_MAX;
        else
          open.push(i);
      }
    }
    vector<int> res;
    int prev_length = 0;
    for (int i = n - 1; i >= 0; --i) {
      if (brackets[i] != INT_MAX) {
        prev_length += (i - brackets[i] + 1);
        if (brackets[i] - 1 < 0 or brackets[brackets[i] - 1] == INT_MAX) {
          res.push_back(prev_length);
          prev_length = 0;
        }
        i = brackets[i];
      }
    }
    if (res.empty()) {
      cout << 0 << ' ' << 1 << '\n';
    } else {
      int max_length = 0;
      for (int i = 0; i < (int)res.size(); ++i)
        max_length = max(max_length, res[i]);
      int count = 0;
      for (int i = 0; i < (int)res.size(); ++i) count += (res[i] == max_length);
      cout << max_length << ' ' << count << '\n';
    }
  }
}