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 n, a[1000000 + 5], f[1000000 + 5], maxl, ans;
string s;
stack<int> stk;
int main() {
  cin >> s;
  n = s.length();
  for (int i = 0; i < n; ++i) {
    if (s[i] == '(')
      stk.push(i);
    else if (s[i] == ')' && !stk.empty())
      a[i] = a[stk.top()] = 1, stk.pop();
  }
  f[0] = a[0];
  for (int i = 1; i < n; ++i)
    f[i] = (a[i] ? f[i - 1] + 1 : 0), maxl = max(maxl, f[i]);
  if (maxl == 0) {
    cout << "0 1" << endl;
    return 0;
  }
  for (int i = 0; i < n; ++i) ans += (f[i] == maxl);
  cout << maxl << " " << ans << endl;
  return 0;
}