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 MAXN = 1e6 + 1;
int f[MAXN];
int s[MAXN];
int e[MAXN];
int main() {
  string S;
  cin >> S;
  int maxi = 0;
  int len = 0;
  int start = 0;
  stack<int> st;
  f[0] = 1;
  s[0] = -1;
  e[0] = -1;
  for (int i = 0; i < S.length(); i++) {
    if (S[i] == '(') {
      st.push(i);
    } else if (st.empty()) {
      e[i] = s[i] = -1;
    } else {
      start = st.top();
      st.pop();
      s[i] = e[i] = start;
      start--;
      if (start >= 0 && S[start] == ')' && s[start] >= 0 && e[start] >= 0) {
        e[i] = e[start];
      }
      len = i - e[i] + 1;
      f[len]++;
      maxi = max(maxi, len);
    }
  }
  cout << maxi << " " << f[maxi] << "\n";
  return 0;
}