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 MX = 1e6 + 100;
int n, d[MX];
string s;
vector<int> v;
int main() {
  cin >> s;
  for (int i = 0; i < s.size(); ++i) {
    if (s[i] == '(')
      v.push_back(i);
    else {
      if (!v.empty()) {
        d[i] = v[v.size() - 1];
        v.pop_back();
      } else
        d[i] = i;
    }
  }
  v.clear();
  for (int i = s.size() - 1; i >= 0; --i) {
    if (s[i] == ')')
      v.push_back(i);
    else {
      if (!v.empty()) {
        d[i] = v[v.size() - 1];
        v.pop_back();
      } else
        d[i] = i;
    }
  }
  int ans = 0, x = 0;
  for (int i = 0; i < s.size(); ++i) {
    if (d[i] != i)
      x++;
    else {
      ans = max(x, ans);
      x = 0;
    }
  }
  ans = max(ans, x);
  if (ans == 0) {
    cout << 0 << " " << 1 << endl;
    return 0;
  }
  cout << ans << " ";
  x = 0;
  int e = 0;
  for (int i = 0; i < s.size(); ++i) {
    if (d[i] != i)
      x++;
    else {
      if (x == ans) e++;
      x = 0;
    }
  }
  if (x == ans) e++;
  cout << e << endl;
}