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 Long_RegularBracketSeq {

	public static void main(String[] args) {
		Scanner sc = new Scanner (System.in);
		String str =  sc.nextLine();
		sc.close();
		int c = 0, freq = 0, max = -1;
		Stack <Integer> stk = new Stack <> ();
		boolean v[] = new boolean[1000001];
		for(int i = 0; i < str.length(); i++) {
			char x = str.charAt(i);
			if(x == '(') 
				stk.push(i);
			else if(!stk.empty()) {
				int z = stk.pop();
				v[i] =true;
				v[z] = true;
				/*c = i - z + 1;
				if(c > max) {
					max = c;
					freq = 1;
				}
				else if(c == max)
					freq++;*/
			}
		}
		for(int i = 0; i < 1000000; i++) {
			c = 0;
			while(v[i] == true) {
				i++;
				c++;
			}
			if(c > max) {
				max = c;
				freq = 1;
			}
			else if(c > 0 && c == max)
				freq++;
		}
		if(max == -1)
			System.out.println("0 1");
		else
			System.out.println(max + " " + freq);
	}
}