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 long long INF = 1e9;
const long long MOD = 1e9 + 7;
string second;
int res = 0;
int idx = 1;
int A[1000005];
int akhir[1000005];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  memset(akhir, -1, sizeof akhir);
  cin >> second;
  int tmp = 0;
  for (int i = 0; i < second.size(); i++) {
    akhir[i] = i;
    if (second[i] == '(') {
      A[++tmp] = i;
    } else if (second[i] == ')') {
      if (tmp == 0) continue;
      akhir[A[tmp]] = i;
      tmp--;
    }
  }
  int now = 0;
  tmp = 0;
  while (now < second.size()) {
    if (akhir[now] == now) {
      tmp = 0;
    } else {
      tmp += (akhir[now] - now + 1);
      if (tmp > res) {
        res = tmp;
        idx = 1;
      } else if (tmp == res) {
        idx++;
      }
    }
    now = akhir[now] + 1;
  }
  cout << res << ' ' << idx << '\n';
}