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;
void solve() {
  string s;
  getline(cin, s);
  int n = s.size(), max_len = 0, cnt_of_max_subs = 1, i = 0, bracketCount = 0;
  map<int, int> a;
  for (char ch : s) {
    if (ch == '(') {
      if (a.find(bracketCount) == a.end()) {
        a[bracketCount] = i;
      }
      ++bracketCount;
    } else {
      --bracketCount;
      if (!a.empty()) {
        auto it = prev(a.end(), 1);
        if (it->first > bracketCount) {
          a.erase(it);
        }
      }
      if (!a.empty()) {
        auto it = prev(a.end(), 1);
        if (it->first == bracketCount) {
          int len = i - it->second + 1;
          if (len > max_len) {
            max_len = len;
            cnt_of_max_subs = 1;
          } else if (len == max_len) {
            ++cnt_of_max_subs;
          }
        }
      }
    }
    ++i;
  }
  cout << max_len << ' ' << cnt_of_max_subs;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int tests = 1;
  while (tests--) solve();
}