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 MAX = 1000000;
int maxi = 0, cant = 1, cont;
int mat[MAX];
stack<int> s;
string curr;
int main() {
  cin >> curr;
  for (int i = 0; i < curr.size(); i++) {
    if (curr[i] == '(')
      s.push(i);
    else if (!s.empty())
      mat[i] = -1, mat[s.top()] = 1, s.pop();
  }
  for (int i = 0; i < curr.size(); i++) {
    if (!mat[i]) {
      if (cont and cont == maxi)
        cant++;
      else if (cont > maxi)
        maxi = cont, cant = 1;
      cont = 0;
      continue;
    }
    cont++;
  }
  if (cont and cont == maxi)
    cant++;
  else if (cont > maxi)
    maxi = cont, cant = 1;
  printf("%d %d\n", maxi, cant);
  return 0;
}