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(0);
  cout.tie(0);
  cin.tie(0);
  string s;
  cin >> s;
  int n = s.size(), ans = 0, cnt = 1, bal = 0;
  stack<int> ss;
  s = '#' + s;
  int a[n + 1];
  memset(a, 0, sizeof a);
  for (int i = 1; i <= n; ++i) {
    if (s[i] == '(')
      ss.push(i), bal++;
    else {
      --bal;
      if (bal < 0) {
        bal = 0;
        while (ss.size()) ss.pop();
        continue;
      }
      int x = ss.top();
      ss.pop();
      a[i] = i - x + 1;
    }
  }
  for (int i = 1; i <= n; ++i) {
    a[i] += a[i - a[i]];
    if (a[i] == ans && ans)
      cnt++;
    else if (a[i] > ans)
      ans = a[i], cnt = 1;
  }
  cout << ans << " " << cnt;
}