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 N = 1e6 + 5;
const int M = 2e5 + 5;
const int OO = 1e9;
const int MOD = 1e9 + 9;
int d[N], c[N], ans, occ = 1;
string second;
stack<int> p;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> second;
  memset(d, -1, sizeof d);
  memset(c, -1, sizeof c);
  for (int i = 0; i < second.size(); ++i) {
    if (second[i] == ')' && !p.empty()) {
      d[i] = p.top();
      p.pop();
      c[i] = (d[i] > 0 && ~(c[d[i] - 1]) ? c[d[i] - 1] : d[i]);
      int cur = i - c[i] + 1;
      if (cur > ans) {
        ans = cur;
        occ = 1;
      } else if (cur == ans)
        ++occ;
    } else if (second[i] == '(')
      p.push(i);
  }
  cout << ans << ' ' << occ << '\n';
  return 0;
}