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 N = 1000005;
int n, lst[N], ans, tot, st[N], tp;
char s[N];
int main() {
  scanf("%s", s + 1);
  n = strlen(s + 1);
  for (int i = 0; i <= n; ++i) lst[i] = i + 1;
  for (int i = 1; i <= n; ++i) {
    if (s[i] == ')') {
      if (tp) {
        int l = st[tp--];
        lst[i] = lst[l - 1];
      }
    } else
      st[++tp] = i;
    ans = max(ans, i - lst[i] + 1);
  }
  if (!ans) {
    puts("0 1");
    return 0;
  }
  for (int i = 1; i <= n; ++i)
    if (i - lst[i] + 1 == ans) tot++;
  printf("%d %d\n", ans, tot);
  return 0;
}