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 c[1000001];
int main(void) {
  stack<int> s;
  int max = 0;
  int cnt = 0;
  int n = 0;
  char cha;
  int i = -1;
  while (cin >> cha) {
    i++;
    if (cha == '(') {
      s.push(i);
      c[i] = -1;
    } else {
      if (s.empty()) {
        c[i] = -1;
      } else {
        int d = s.top();
        s.pop();
        if (d - 1 >= 0 && c[d - 1] >= 0) {
          c[i] = c[d - 1];
        } else {
          c[i] = d;
        }
        if (i - c[i] + 1 > max) {
          max = i - c[i] + 1;
          cnt = 1;
        } else if (i - c[i] + 1 == max) {
          cnt++;
        }
      }
    }
  }
  if (max > 0)
    cout << max << " " << cnt << endl;
  else
    cout << 0 << " " << 1 << endl;
  return 0;
}