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 S[1000001];
int dp[1000001];
int ans, ansi;
int main() {
  scanf("%s", S);
  int N = strlen(S);
  for (int i = 1; i < N; i++) {
    if (S[i] == ')') {
      if (dp[i - 1] > 0) {
        if (S[i - dp[i - 1] - 1] == '(') {
          dp[i] = dp[i - 1] + 2;
          if (dp[i - dp[i - 1] - 2] > 0) dp[i] += dp[i - dp[i - 1] - 2];
        }
      } else {
        if (S[i - 1] == '(') {
          dp[i] = 2;
          dp[i] += dp[i - 2];
        }
      }
    }
  }
  for (int i = 0; i < N; i++) {
    if (dp[i] == ans) {
      ansi++;
    } else if (dp[i] > ans) {
      ans = dp[i];
      ansi = 1;
    }
  }
  if (ans == 0)
    cout << 0 << ' ' << 1 << endl;
  else
    cout << ans << ' ' << ansi << endl;
}