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;
stack<int> sk;
bool trash[1000 * 1000 + 10];
int main() {
  string s;
  cin >> s;
  int ans = -1, cnt = 1, tmp = 0;
  for (int i = 0; i < (int)s.size(); i++) {
    if (s[i] == '(')
      sk.push(i);
    else if (!sk.empty())
      sk.pop();
    else
      trash[i] = true;
  }
  while (!sk.empty()) {
    trash[sk.top()] = true;
    sk.pop();
  }
  for (int i = 0; i < (int)s.size(); i++) {
    if (trash[i])
      tmp = 0;
    else
      tmp++;
    if (tmp > ans) {
      ans = tmp;
      cnt = 1;
    } else if (tmp == ans)
      cnt++;
  }
  if (ans == 0) cnt = 1;
  cout << ans << " " << cnt << endl;
  return 0;
}