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;
string s;
int poz[1000100], maxx;
stack<int> S;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> s;
  for (int i = 0; i < s.size(); i++) poz[i] = -2;
  for (int i = 0; i < s.size(); i++) {
    if (s[i] == '(')
      S.push(i);
    else {
      if (S.empty()) {
        poz[i] = -1;
        continue;
      }
      poz[S.top()] = i;
      poz[i] = S.top();
      S.pop();
    }
  }
  for (int i = 0; i < s.size(); i++)
    if (poz[i] == -2) poz[i] = -1;
  for (int i = 0; i < s.size(); i++) {
    int cnt = 0;
    while (i < s.size() && poz[i] != -1) ++cnt, ++i;
    maxx = max(cnt, maxx);
  }
  int rs = 0;
  for (int i = 0; i < s.size(); i++) {
    int cnt = 0;
    while (i < s.size() && poz[i] != -1) ++cnt, ++i;
    rs += (maxx == cnt);
  }
  if (!maxx) return cout << 0 << " " << 1, 0;
  cout << maxx << " " << rs;
  cerr << "Fucking time elapsed: " << clock() * 1000.0 / CLOCKS_PER_SEC << " ms"
       << '\n';
  return 0;
}