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 JAVA):

Solution:
/**
 * Created with IntelliJ IDEA.
 * User: yuantian
 * Date: 3/27/13
 * Time: 11:41 PM
 * Copyright (c) 2013 All Right Reserved, http://github.com/tyuan73
 */

import java.util.*;

public class LongestRegularBracketSequence5C {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);

        String b = in.next();
        int n = b.length();
        int[] dp = new int[n+1];

        for(int i = n-2; i >= 0; i--) {
            if(b.charAt(i) == '(') {
                int next = i+dp[i+1]+1;
                if(next < n && b.charAt(next) == ')') {
                    dp[i] = dp[i+1] + 2 + dp[next+1];
                }
            }
        }

        int max = 0, count = 1;
        for(int i = 0; i < n; i++) {
            if(dp[i] > max) {
                max = dp[i];
                count = 1;
            } else if(dp[i] == max && max != 0) {
                count++;
            }
        }

        System.out.println(max + " " + count);
    }
}