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() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  string s;
  cin >> s;
  int n = s.size();
  int dp[n];
  dp[0] = 0;
  for (int i = 1; i < n; i++) {
    if (s[i] == '(')
      dp[i] = 0;
    else if (i - dp[i - 1] - 1 < 0)
      dp[i] = 0;
    else if (s[i - dp[i - 1] - 1] == ')')
      dp[i] = 0;
    else if (i - dp[i - 1] - 2 < 0)
      dp[i] = dp[i - 1] + 2;
    else {
      dp[i] = 2 + dp[i - 1] + dp[i - dp[i - 1] - 2];
    }
  }
  int cont = 1;
  int max = dp[0];
  for (int i = 0; i < n; i++) {
    if (dp[i] == max && dp[i] != 0)
      cont++;
    else if (dp[i] > max) {
      max = dp[i];
      cont = 1;
    }
  }
  cout << max << " " << cont;
  return 0;
}