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.*;
import java.util.*;

public class Main {

	public static void main(String[] args) throws IOException{
		BufferedReader b = new BufferedReader(new InputStreamReader(System.in));
		char[] s = b.readLine().toCharArray();
		int max = 0;
		int n = 1;
		int j = -1;
		int bal = 0;
		for (int i = 0; i < s.length; i++) {
			bal += (s[i] == '(') ? 1 : -1;
			if(bal == 0){
				if(i - j == max){
					n++;
				}
				if(i - j > max){
					max = i - j;
					n = 1;
				}
			}
			if(bal < 0){
				j = i;
				bal = 0;
			}
		}
		if(bal != 0){
			j = s.length;
			bal = 0;
			for (int i = s.length - 1; i >= 0; i--) {
				bal += (s[i] == ')') ? 1 : -1;
				if(bal == 0){
					if(j - i == max){
						n++;
					}
					if(j - i > max){
						max = j - i;
						n = 1;
					}
				}
				if(bal < 0){
					j = i;
					bal = 0;
				}
			}
		}
		System.out.println(max + " " + n);
	}
}