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;
int main(void) {
  string input;
  while (cin >> input) {
    stack<int> chk;
    int single[1000001] = {(0)}, maxi = 0, rept[1000001] = {(0)};
    for (int i = 0; i < input.size(); i++) {
      if (input[i] == '(') {
        chk.push(i);
      } else {
        if (chk.empty()) {
          single[i] = -1;
        } else {
          int indx = chk.top();
          chk.pop();
          single[i] = indx;
          if (indx > 0 && input[indx - 1] == ')' && single[indx - 1] >= 0) {
            single[i] = single[indx - 1];
          }
          int length = i - single[i] + 1;
          rept[length]++;
          maxi = max(length, maxi);
        }
      }
    }
    rept[0] = 1;
    cout << maxi << " " << rept[maxi] << endl;
  }
}