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:
import java.util.*;
public class Main
{
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		String str= sc.nextLine();
        int Mseq=0,Cseq=0,seq=0;
		Stack<Character> t=new Stack<Character>();
        Stack<Integer> res= new Stack<Integer> ();
        res.push(0);
		for(int i=0;i<str.length();i++)
		{
		    char ch= str.charAt(i);
		    if(ch=='('){
		    t.push(ch);
		    res.push(0);
		    }
		    else if(ch==')'&&!t.isEmpty())
		    {
		    t.pop();
		        int a=res.pop();
		        int b=res.pop();
		        int val=a+b+2;
		        res.push(val);
		        if(Mseq<val)
		            Mseq=val;
		    }
		    else{
		     seq=0;
		    res.push(0);
		        
		    }
		}
		while(!res.isEmpty())
		{   
		    if(Mseq==res.pop())
		    Cseq++;
		}
		if(Mseq==0)
		System.out.println("0 1");
		else
		System.out.println(Mseq+" "+Cseq);

	}
}