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.*;
import java.io.*;
public class CodeForces5C {
	public static void main(String[] args) throws IOException{
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		PrintWriter w = new PrintWriter(System.out);
		String str = br.readLine();
		int n = str.length() - 1;
		Stack<Integer> stk = new Stack<Integer>();
		int[] dp = new int[n + 1];
		int max = 0;
		for(int i = 0; i <= n; i++) {
			if(str.charAt(i) == '(')
				stk.push(i);
	
			else if(!stk.isEmpty()) {			
				dp[i] = (stk.peek()==0 ? i-stk.peek()+1 : i-stk.peek()+1+dp[stk.peek()-1]);
				max = Math.max(dp[i], max);
				stk.pop();
			}
		}
		if(max == 0) {
			System.out.println(0 + " " + 1);
		}
		else {
			int count = 0;
			for(int i = 0; i <= n; i++) {
				if(dp[i] == max) {
					count++;
				}
//			System.out.println(dp[i]);
			}
			System.out.println(max + " " + count);
		}
	}
}