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 M4 = 1e4 + 100, M5 = 1e5 + 100, M6 = 1e6 + 100, M7 = 1e7 + 100,
          M8 = 1e8 + 100, M9 = 1e9 + 100;
int cnt[M6];
stack<int> st;
int ans, q;
void clr() {
  while ((int)(st).size()) st.pop();
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  string s;
  cin >> s;
  int t = 0;
  for (int i = 0; i < (int)(s).size(); i++) {
    char c = s[i];
    if (c == '(')
      st.push(i);
    else if ((int)(st).size()) {
      cnt[i] = i - st.top() + 1;
      if (st.top() > 0) cnt[i] += cnt[st.top() - 1];
      st.pop();
    }
  }
  for (int i = 0; i < (int)(s).size(); i++) ans = max(ans, cnt[i]);
  for (int i = 0; i < (int)(s).size(); i++)
    if (cnt[i] == ans) q++;
  cout << ans << ' ' << (ans == 0 ? 1 : q);
  return 0;
}