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 maxn = 1e6 + 10;
int dp[maxn];
long long gcd(long long a, long long b) { return !b ? a : gcd(b, a % b); }
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  string s;
  cin >> s;
  stack<int> k;
  for (int i = 0; i < s.size(); i++) {
    if (s[i] == '(')
      k.push(i);
    else {
      if (!k.empty()) {
        dp[i] = i - k.top() + 1;
        if (k.top() > 0) dp[i] += dp[k.top() - 1];
        k.pop();
      }
    }
  }
  int ans1 = 0, ans2 = 1;
  for (int i = 0; i < s.size(); i++) {
    if (dp[i] > ans1) {
      ans1 = dp[i];
      ans2 = 1;
    } else if (dp[i] == ans1)
      ans2++;
  }
  if (ans1 == 0)
    printf("0 1\n");
  else
    cout << ans1 << " " << ans2 << endl;
  return 0;
}