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;
long long a[1000006], s[1000006];
int main() {
  ios_base::sync_with_stdio(false);
  long long n, i, j, k;
  string st;
  cin >> st;
  long long ln = st.size();
  stack<long long> stk;
  for (i = 0; i < ln; i++) {
    if (st[i] == '(') {
      stk.push(i);
    } else {
      if (stk.empty()) {
        s[i] = -1;
      }
      if (!stk.empty()) {
        long long idx = stk.top();
        stk.pop();
        if (idx > 0 && st[idx - 1] == ')' && s[idx - 1] >= 0)
          s[i] = s[idx - 1];
        else
          s[i] = idx;
        long long xx = i - s[i] + 1;
        a[xx]++;
      }
    }
  }
  for (i = 1000005; i >= 0; i--) {
    if (a[i] > 0) {
      cout << i << " " << a[i] << endl;
      return 0;
    }
  }
  cout << 0 << " " << 1 << endl;
  return 0;
}