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() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int test = 1;
  while (test--) {
    string s;
    cin >> s;
    int n = s.length();
    int dp[n];
    memset(dp, -1, sizeof dp);
    stack<int> ss;
    for (int i = 0; i < n; i++) {
      if (s[i] == '(') {
        ss.push(i);
      } else {
        if (ss.size()) {
          dp[i] = ss.top();
          ss.pop();
        }
      }
    }
    int ans = 0, cnt = 1;
    for (int i = (n - 1); i >= 0; i--) {
      if (dp[i] == -1) {
        continue;
      }
      int j = dp[i] - 1;
      while (j >= 0 && dp[j] != -1) {
        j = dp[j] - 1;
      }
      if (ans < (i - j)) {
        ans = i - j;
        cnt = 1;
      } else if (ans == (i - j)) {
        cnt++;
      }
      i = j + 1;
    }
    cout << ans << " " << cnt << endl;
  }
  return 0;
}