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 NN = (int)1e6 + 6;
int L[NN];
int R[NN];
char s[NN];
int main() {
  scanf("%s", s);
  int n = strlen(s);
  int balance = 0;
  for (int i = 0; i < n; i++) {
    if (s[i] == '(')
      balance++;
    else if (s[i] == ')')
      balance--;
    if (balance == -1) s[i] = '?', balance = 0;
  }
  balance = 0;
  for (int i = n - 1; i >= 0; i--) {
    if (s[i] == ')')
      balance++;
    else if (s[i] == '(')
      balance--;
    if (balance == -1) s[i] = '?', balance = 0;
  }
  int ans = 0, cnt = 1;
  int idx = -1;
  for (int i = 0; i < n; i++) {
    if (s[i] == '?')
      idx = i;
    else {
      if (i - idx > ans) {
        ans = i - idx, cnt = 1;
      } else if (i - idx == ans)
        cnt++;
    }
  }
  cout << ans << ' ' << cnt << endl;
  return 0;
}