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 str[1000005];
int d[1000005];
int main() {
  stack<int> s;
  memset(d, 0, sizeof(d));
  scanf("%s", str + 1);
  int len = strlen(str + 1);
  for (int i = 1; i <= len; i++) {
    if (str[i] == '(')
      s.push(i);
    else {
      if (!s.empty()) {
        int k = s.top();
        s.pop();
        d[i] = d[k - 1] + i - k + 1;
      }
    }
  }
  int maxlen = 0, cnt = 1;
  for (int i = 1; i <= len; i++)
    if (str[i] == ')') {
      if (d[i] == maxlen)
        cnt++;
      else if (d[i] > maxlen) {
        maxlen = d[i];
        cnt = 1;
      }
    }
  if (maxlen)
    cout << maxlen << " " << cnt << '\n';
  else
    cout << "0 1\n";
  return 0;
}