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 C
{
	public static void main( String[] args )
	{
		Scanner in = new Scanner( System.in );
		String str = in.next();
		int n = str.length(), maxi = 0, count = 0;
		char[] ch = str.toCharArray();
		int[] f = new int[n + 1];
		Stack<Integer> st = new Stack<>();
		for(int i = 0; i < n; i++)
		{
			if(ch[i] == '(') st.push(i);
			else
			{
				if(!st.isEmpty())
				{
					// int pop = st.pop();
					f[i + 1] = 2 + f[st.pop()] + f[i];
				}
			}
			maxi = Math.max(maxi, f[i + 1]);
		}
		
		for(int v : f) if(maxi == v) count++;
		if(maxi == 0) System.out.printf("0 1");
		else System.out.printf("%d %d", maxi, count);
	}
}