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;
stack<long long int> d, re;
long long int k = 0, e = -1, m = 0, a[30000000];
int main() {
  cin >> s;
  d.push(7);
  a[0] = 0;
  a[1] = 0;
  for (int i = 0; i < s.size(); i++) {
    if (s[i] == '(') {
      d.push(1);
      if (s[i - 1] == ')')
        a[i + 2] = a[i + 1];
      else
        a[i + 2] = 0;
    }
    if (s[i] == ')') {
      if (d.top() == 1) {
        e = a[i + 1];
        e += 2 + a[i + 2 - e - 1];
        a[i + 2] = e;
        if (e > m) {
          k = 1;
          m = e;
        } else if (e == m)
          k++;
        d.pop();
      } else {
        d.push(0);
        a[i + 2] = 0;
      }
    }
  }
  if (k == 0) {
    cout << 0 << " " << 1;
    return 0;
  }
  cout << m << ' ' << k;
  return 0;
}