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>
int i, top, max, nmax, cnt, st[1000001];
int main() {
  char str[1000000];
  scanf("%s", str);
  for (; str[i]; ++i) {
    if (str[i] == '(')
      st[top++] = i;
    else {
      if (top > 0) {
        str[i] = '.';
        str[st[--top]] = '.';
      }
    }
  }
  max = nmax = cnt = 0;
  for (i = 0; str[i]; ++i) {
    if (str[i] == '.')
      ++cnt;
    else {
      if (cnt > max) {
        max = cnt;
        nmax = 1;
      } else if (max == cnt)
        ++nmax;
      cnt = 0;
    }
  }
  if (cnt > max) {
    max = cnt;
    nmax = 1;
  } else if (cnt == max)
    ++nmax;
  if (max == 0) nmax = 1;
  printf("%d %d\n", max, nmax);
}