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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Stack;

public class MainA {
	public static void main(String[] args) throws IOException {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		char[] s = br.readLine().toCharArray();
		int longest = 0;
		int count = 1;
		int[] l = new int[s.length];
		Stack<Integer> p = new Stack<>();
		for (int i = 0; i < s.length; i++) {
			if (s[i] == ')') {
				if (p.isEmpty()) {
					l[i] = -1;
				} else {
					int x = p.pop();
					l[i] = i - x + 1;
					if (x > 0 && s[x - 1] == ')' && l[x - 1] > 0) {
						l[i] += l[x - 1];
					}
				}
			} else {
				l[i] = -1;
				p.push(i);
			}
		}

		for (int i = 0; i < l.length; i++) {
			if (l[i] > longest) {
				longest = l[i];
				count = 1;
			} else if (l[i] == longest) {
				count++;
			}
		}

		System.out.println(longest + " " + count);
	}
}