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 CF5C {
	public static void main(String[] args){
		Scanner br = new Scanner(System.in);
		String line = br.next();
		Stack<Integer> st = new Stack<Integer>();
		Stack<Integer> ends = new Stack<Integer>();
		int max = 0;
		int count = 1;
		for(int i = 0;i<line.length();i++){
			if(line.charAt(i) == ')'){
				if(st.size() > 0){
					int cur = st.pop();
					while(ends.size() > 0 && ends.peek() > cur){
						ends.pop();
						ends.pop();
					}
					
					int len = (i-cur+1);
					if(ends.size() > 0 && ends.peek() == cur-1){
						int temp = ends.pop();
						len+=ends.peek();
						ends.push(temp);
					}
					if(len > max){
						max = len;
						count = 1;
					}
					else if(len == max){
						count++;
					}
					ends.push(len);
					ends.push(i);
					
				}
				else{
					ends.clear();
				}
			}
			else{
				st.push(i);
			}
		}
		System.out.println(max+" "+count);
	}
}