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;
char str[1000005];
stack<int> s;
int dp[1000005];
int main() {
  while (scanf("%s", str) != -1) {
    int len = strlen(str);
    int i, j;
    while (!s.empty()) {
      s.pop();
    }
    int cnt = 0;
    int maxm = 0;
    s.push(-1);
    for (i = 0; i < len; i++) {
      if (str[i] == '(') {
        s.push(i);
      } else {
        if (s.size() <= 1) {
          s.pop();
          s.push(i);
        } else {
          s.pop();
          int m = i - s.top();
          if (m > maxm) {
            maxm = m;
            cnt = 1;
          } else if (m == maxm)
            cnt++;
        }
      }
    }
    if (maxm)
      cout << maxm << " " << cnt << endl;
    else
      cout << 0 << " " << 1 << endl;
  }
  return 0;
}