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() {
  string s;
  cin >> s;
  int k = s.size();
  stack<int> st;
  int dp[k];
  memset(dp, -1, sizeof dp);
  for (int i = 0; i < k; i++) {
    if (s[i] == '(') st.push(i);
    if (s[i] == ')') {
      if (!st.empty()) {
        int p = st.top();
        if (p - 1 >= 0 && s[p - 1] == ')' && dp[p - 1] != -1)
          dp[i] = dp[p - 1];
        else
          dp[i] = p;
        st.pop();
      }
    }
  }
  int maxi = 0;
  for (int i = 0; i < k; i++) {
    if (dp[i] != -1) {
      maxi = max(maxi, i - dp[i] + 1);
      dp[i] = i - dp[i] + 1;
    }
  }
  if (maxi == 0)
    cout << "0 1\n";
  else {
    int c = 0;
    for (int i = 0; i < k; i++) {
      if (dp[i] == maxi) c++;
    }
    cout << maxi << " " << c << "\n";
  }
}