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 MAXN = 1000005;
int num[MAXN];
int main() {
  string str;
  cin >> str;
  stack<int> stk;
  int count = 0;
  for (int i = 0; i < str.size(); ++i) {
    if (str[i] == '(') {
      stk.push(i);
    } else if (str[i] == ')' && !stk.empty()) {
      num[stk.top()] += i - stk.top() + 1;
      if (i + 1 < str.size() && str[i + 1] == '(') num[i + 1] += num[stk.top()];
      stk.pop();
    }
  }
  while (!stk.empty()) {
    num[stk.top()] = 0;
    stk.pop();
  }
  int max = 0, times = 0;
  for (int i = 0; i < str.size(); ++i) {
    if (str[i] == ')') continue;
    if (num[i] > max) {
      max = num[i];
      times = 1;
    } else if (num[i] == max) {
      ++times;
    }
  }
  if (max == 0)
    printf("0 1");
  else
    printf("%d %d", max, times);
  return 0;
}