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;
string s;
int n, open[1000000 + 3], dp[1000000 + 3];
int main() {
  cin >> s;
  n = (int)s.length();
  memset(open, -1, sizeof(open));
  memset(dp, 0, sizeof(dp));
  stack<int> st;
  for (int i = 0; i < n; i++) {
    if (s[i] == '(')
      st.push(i);
    else {
      if (st.empty() || s[st.top()] != '(')
        continue;
      else {
        open[i] = st.top();
        st.pop();
      }
    }
  }
  for (int i = 0; i < n; i++) {
    if (s[i] == '(' || open[i] == -1) continue;
    int prv = open[i] - 1;
    dp[i] = (i - prv) + (prv == 0 ? 0 : dp[prv]);
  }
  int best = 0;
  int cnt = 1;
  for (int i = 0; i < n; i++) {
    if (dp[i] > best) {
      best = dp[i];
      cnt = 1;
    } else if (dp[i] == best) {
      cnt += 1;
    }
  }
  if (best == 0)
    cout << "0 1\n";
  else
    cout << best << " " << cnt << endl;
}