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>
#pragma comment(linker, "/STACK:64000000")
using namespace std;
int main() {
  stack<pair<int, int> > st;
  char buf[1000000 + 10];
  scanf("%s", buf);
  string s = string(buf);
  int m = 0;
  int n = 1;
  int sum = 0;
  vector<int> start(2 * 1000000 + 10, -1);
  vector<int> num(1000000 + 10, 0);
  for (int i = 0; i < s.size(); ++i) {
    if (s[i] == '(') {
      if (start[1000000 + sum] == -1) start[1000000 + sum] = i;
      ++sum;
    } else {
      --sum;
      if (start[1000000 + sum] != -1) {
        ++num[1 + i - start[1000000 + sum]];
      }
      start[1000000 + sum + 1] = -1;
    }
  }
  num[0] = 1;
  for (int m = num.size() - 1; m >= 0; --m) {
    if (num[m] > 0) {
      cout << m << " " << num[m] << "\n";
      return 0;
    }
  }
  return 0;
}