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);
  cin.tie(0);
  cout.tie(0);
  string s;
  cin >> s;
  long long n = s.size();
  stack<int> stk;
  long long len = 0, mx = 0;
  map<long long, long long> f, st;
  f[0] = 1;
  for (int i = 0; i <= n - 1; i++) {
    int flag = 0;
    if (s[i] == ')') {
      if (stk.size() == 0) {
        continue;
      }
      int w = stk.top();
      stk.pop();
      len = (i - w + 1);
      st[w] = len;
    } else {
      stk.push(i);
    }
  }
  for (int i = n - 1; i >= 0; i--) {
    st[i] = st[i] + st[i + st[i]];
    mx = max(mx, st[i]);
    if (st[i]) f[st[i]]++;
  }
  cout << (mx) << " " << f[mx] << endl;
}