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[1000010];
int a[1000010];
int main(void) {
  int i, j, k, n, m, l, x, y;
  memset(a, 0, sizeof(a));
  scanf("%s", s);
  l = strlen(s);
  x = y = 0;
  k = 0;
  n = 0;
  for (i = 0; i < l; i++) {
    if (s[i] == '(')
      x++;
    else {
      if (x > 0) a[i] = 1;
      x--;
    }
    if (x < 0) {
      x = 0;
    }
  }
  x = 0;
  for (i = l - 1; i >= 0; i--) {
    if (s[i] == ')')
      x++;
    else {
      if (x > 0) a[i] = 1;
      x--;
    }
    if (x < 0) {
      x = 0;
    }
  }
  k = 0;
  j = 0;
  for (i = 0; i < l; i++) {
    if (a[i])
      k++;
    else
      k = 0;
    if (k > n) {
      n = k;
      j = 1;
    } else if (k == n) {
      j++;
    }
  }
  if (n == 0) j = 1;
  printf("%d %d\n", n, j);
  return 0;
}