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 Brackets{
	public static void main(String[] args){
		Scanner sc = new Scanner(System.in);
		Stack<Integer> stack = new Stack<Integer>();
		String input = sc.next();
		int len = input.length();
		int[] d = new int[len];
		int[] c = new int[len];
		int current = 0;
		int longest = 0;
		int count = 0;
		for(int i=0;i<len;i++){
			if(input.charAt(i) == '('){
				stack.push(i);
				d[i] = c[i] = -1;
			}
			else{
				if(!stack.isEmpty()){
					d[i] = stack.pop();
					if((d[i]-1) >= 0 && input.charAt(d[i]-1) == ')' && c[d[i]-1] != -1){
						c[i] = c[d[i]-1];
						current = d[i] - 1 - c[d[i]-1] + 1 + i - d[i] + 1;
						if(current > longest){
							longest = current;
							count = 1;
						}
						else if(current == longest) count++;
					}else{
						current = i - d[i] + 1;
						if(current > longest){
							longest = current;
							count = 1;
						}
						else if(current == longest) count++;
						c[i] = d[i];
					}
				}else d[i] = c[i] = -1;
			}
		}
		if(longest == 0 && count == 0) System.out.println(0 + " " + 1);
		else System.out.println(longest + " " + count);
	}
}