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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Stack;


public class Main {
	
	
	public static String areParenthesis(String[] a) {
		String msg="";
		Stack<Integer> st = new Stack<Integer>();
		st.push(-1);
		int tama�oMaximoSubCad=0;
		int totalCad=0;
		for (int i = 0; i < a.length; i++) {
			
			if(a[i].equals("(")){
				st.push(i);
			}else if(a[i].equals(")")) {
				st.pop();
				if(!st.isEmpty()) {
					int totalCad2= i-st.peek();
					if(totalCad2>tama�oMaximoSubCad) {
						tama�oMaximoSubCad=totalCad2;
						totalCad=1;
					}else if(totalCad2==tama�oMaximoSubCad) {
						totalCad++;
					}
				}else {
					st.push(i);
				}
			}
		}
		if(totalCad==0) {
			msg="0 1";
		}else {
			msg=tama�oMaximoSubCad+" "+totalCad;
		}
		
		return msg;
		
	}
	public static void main(String[] args) throws IOException {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		String[] line=br.readLine().split("");
		System.out.println(areParenthesis(line)); 
			
	}

}