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;
int ok[1000001];
int main() {
  stack<pair<int, char> > S;
  int cnt = 0;
  char c;
  int i = 0;
  while (cin >> c) {
    if (S.size() == 0 || c == '(') {
      S.push(make_pair(i, c));
    } else if (S.size() != 0 && S.top().second == '(' && c == ')') {
      ok[S.top().first] = ok[i] = true;
      S.pop();
    }
    i++;
  }
  int len = 0, ans = 0, ans2 = 0;
  for (int i = 0; i < 1000001; i++) {
    if (ok[i]) {
      len++;
    } else {
      ans = max(len, ans);
      len = 0;
    }
  }
  for (int i = 0; i < 1000001; i++) {
    if (ok[i]) {
      len++;
    } else {
      if (ans == len) ans2++;
      len = 0;
    }
  }
  if (ans == 0)
    cout << 0 << " " << 1 << endl;
  else
    cout << ans << " " << ans2 << endl;
}