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;
struct p {
  int u, v, w;
  p() {}
  p(int u, int v, int w) : u(u), v(v), w(w) {}
  bool operator<(const p &right) { return this->w < right.w; }
};
int n;
string s1, s2;
map<pair<char, char>, int> M;
map<char, int> S;
int l, r;
int main() {
  vector<int> ans;
  string s;
  cin >> s;
  vector<int> dp(s.size() + 1, 0);
  s = s + "(";
  for (int i = s.size() - 3; i >= 0; --i)
    if (s[i] == '(' && s[i + dp[i + 1] + 1] == ')')
      dp[i] = dp[i + 1] + dp[i + dp[i + 1] + 2] + 2;
  int maxn = *max_element(dp.begin(), dp.end());
  if (maxn == 0)
    cout << "0 1" << endl;
  else {
    cout << maxn << " ";
    cout << count(dp.begin(), dp.end(), maxn);
  }
}