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 Parenthesis {

	public static void main(String[] args) throws IOException {
		BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
		String s = in.readLine();
		Stack<Integer> stk = new Stack<Integer>();
		stk.push(-1);
		int count = 0;
		int current = 0;
		int max = 0;
		for (int i = 0; i < s.length(); i++) {
			char c = s.charAt(i);
			if(c == '('){
				stk.push(i);
			}else{
				stk.pop();
				if(!stk.isEmpty()){
					int aux = stk.peek();
					current = (i - aux);
					if(current > max){
						max = current;
						count = 1;
					}else if(current > 0 && current == max){
						count++;
					}
				}else{
					stk.push(i);
				}
			}
		}
		if(count > 0)
			System.out.println(max + " " + count);
		else
			System.out.println("0 1");
	}

}