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 str, int len, int& sublen, int& subnum) {
  int cnt = 0;
  int start = 0;
  for (int i = 0; i < len; i++) {
    cnt += (str[i] == '(' ? 1 : 0);
    cnt -= (str[i] == ')' ? 1 : 0);
    if (cnt == 0) {
      if (i - start + 1 > sublen) {
        sublen = i - start + 1;
        subnum = 1;
      } else if (i - start + 1 == sublen)
        subnum++;
    } else if (cnt < 0) {
      start = i + 1;
      cnt = 0;
    }
  }
}
int main() {
  string str, restr;
  getline(cin, str);
  restr = str;
  int len = str.length();
  int j = 0;
  for (int i = len - 1; i >= 0; i--) restr[j++] = (str[i] == '(' ? ')' : '(');
  int sublen1 = 0, subnum1 = 0, sublen2 = 0, subnum2 = 0;
  solve(str, len, sublen1, subnum1);
  solve(restr, len, sublen2, subnum2);
  if (sublen1 == 0 && sublen2 == 0)
    cout << 0 << " " << 1 << endl;
  else if (sublen2 > sublen1)
    cout << sublen2 << " " << subnum2 << endl;
  else
    cout << sublen1 << " " << subnum1 << endl;
  return 0;
}