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 dp[1000005];
stack<int> left_bracket;
int main() {
  string str;
  cin >> str;
  int i, cnt = 0, max = -1, slen;
  slen = str.length();
  for (i = 0; i < slen; i++) {
    if (str[i] == '(')
      left_bracket.push(i);
    else {
      if (!left_bracket.empty()) {
        int t = left_bracket.top();
        left_bracket.pop();
        dp[i] = dp[t - 1] + i - t + 1;
        if (dp[i] > max) {
          max = dp[i];
          cnt = 1;
        } else if (dp[i] == max)
          cnt++;
      }
    }
  }
  if (cnt == 0)
    printf("0 1");
  else
    printf("%d %d", max, cnt);
  return 0;
}