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;
const int MAX = 10e5 + 1;
int dp[MAX];
int main() {
  stack<int> s;
  string ss;
  cin >> ss;
  for (int i = 0; i < ss.length(); i++) {
    if (ss[i] == '(') {
      s.push(i);
    } else if (!s.empty()) {
      int t = s.top();
      s.pop();
      if (t) {
        dp[i] = dp[t - 1] + (i - t + 1);
      } else {
        dp[i] = dp[0] + (i - t + 1);
      }
    }
  }
  int max = *max_element(dp, dp + ss.length());
  if (max == 0) {
    cout << "0 1" << endl;
    return 0;
  } else {
    int c = count(dp, dp + ss.length(), max);
    cout << max << ' ' << c << endl;
  }
  return 0;
}