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 int MAXN = (int)1e6 + 7;
const int INF = (int)2e9 + 7;
int cnt[MAXN];
string s;
vector<int> pos;
vector<pair<int, int>> all;
int main() {
  cin >> s;
  for (int i = 0; i < s.size(); i++) {
    if (s[i] == '(')
      pos.push_back(i);
    else {
      if (pos.empty() || s[pos.back()] != '(')
        pos.clear();
      else {
        while (!all.empty() && pos.back() <= all.back().first &&
               all.back().second <= i)
          all.pop_back();
        if (!all.empty() && all.back().second + 1 == pos.back())
          all[all.size() - 1].second = i;
        else
          all.push_back(make_pair(pos.back(), i));
        pos.pop_back();
      }
    }
  }
  for (auto i : all) cnt[i.second - i.first + 1]++;
  cnt[0]++;
  for (int i = s.size(); i >= 0; i--) {
    if (cnt[i]) {
      printf("%d %d", i, cnt[i]);
      return 0;
    }
  }
  return 0;
}