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 main() {
  string str;
  cin >> str;
  long long n = str.length();
  stack<long long> st;
  long long d[n + 1];
  long long c[n + 1];
  for (long long i = 0; i < n; i++) {
    d[i] = -1;
    c[i] = -1;
  }
  for (long long i = 0; i < n; i++) {
    if (str[i] == '(')
      st.push(i);
    else {
      if (st.empty()) {
        d[i] = c[i] = -1;
        continue;
      }
      d[i] = st.top();
      st.pop();
      if ((d[i] - 1) > 0 && str[d[i] - 1] == ')' && c[d[i] - 1] != -1)
        c[i] = c[d[i] - 1];
      else
        c[i] = d[i];
    }
  }
  long long ans = 0;
  long long cnt = 1;
  for (long long i = 0; i < n; i++) {
    if (c[i] != -1) {
      if ((i - c[i] + 1) > ans) {
        ans = i - c[i] + 1;
        cnt = 1;
      } else if ((i - c[i] + 1) == ans)
        cnt++;
    }
  }
  cout << ans << " " << cnt;
  return 0;
}