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 void main(String[]args) throws IOException {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		secuence(br.readLine());
		br.close();
		
	}
	public static void secuence(String c){
		Stack<Integer> s = new Stack<>();
		s.push(-1);
		int tamMax = 0;
		int totalCadenas = 0;
		for(int i = 0; i< c.length(); i++) {
			if(c.charAt(i) == '(') {
				s.push(i);
			}else {
				s.pop();
				if(s.isEmpty() == false) {
					int l = i - s.peek();
					if(l > tamMax) {
						tamMax = l;
						totalCadenas = 1;
					}else if(l == tamMax){
						totalCadenas++;
					}
				}else {
					s.push(i);
				}
			}
		}
		if(totalCadenas == 0) {
			System.out.println("0 1");
		}else {
			System.out.println(tamMax + " " + totalCadenas);
		}
	}
}