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;
char s[1010000];
int len[1010000];
int main() {
  scanf("%s", s);
  stack<int> st;
  int n = 0;
  for (; s[n] != 0; ++n) {
    if (s[n] == '(') {
      st.push(n);
    } else {
      if (!st.empty()) {
        int last = st.top();
        st.pop();
        if (last == 0)
          len[n] = n + 1;
        else
          len[n] = len[last - 1] + n - last + 1;
      }
    }
  }
  int ansl = 0, ansc = 1;
  for (int i = 0; i < n; ++i) {
    if (len[i] == ansl) {
      ansc++;
    } else if (len[i] > ansl) {
      ansl = len[i];
      ansc = 1;
    }
  }
  if (ansl == 0) ansc = 1;
  printf("%d %d\n", ansl, ansc);
  return 0;
}