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 pos[1000000 + 5];
char str[1000000 + 5];
int main() {
  scanf("%s", str + 1);
  int N = strlen(str + 1), Ans = 0, cnt = 0;
  for (int i = 1; i <= N; i++) {
    int l = i, r = i + 1;
    if (str[l] == '(' && str[r] == ')') {
      while (l >= 1 && r <= N) {
        l--, r++;
        if (str[l] == ')' && pos[l]) l = pos[l] - 1;
        if (str[l] != '(' || str[r] != ')') break;
      }
      pos[r - 1] = l + 1;
    }
    if (pos[i]) Ans = max(Ans, i - pos[i] + 1);
  }
  if (!Ans) {
    puts("0 1");
    return 0;
  }
  for (int i = 1; i <= N; i++)
    if (pos[i]) cnt += (i - pos[i] + 1 == Ans);
  printf("%d %d", Ans, cnt);
}