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.io.PrintWriter;
import java.util.Stack;

public class Main {

	public static void main(String[] args) throws IOException {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		PrintWriter out = new PrintWriter(System.out);
		String exp = br.readLine();
		int[] dist = new int[exp.length()];
		Stack<Integer> valid = new Stack<>();
		int counter = 0;
		int max = 0;
		int occ = 0;
		for (int i = 0; i < exp.length(); i++) {
			char current = exp.charAt(i);
			if (current == '(') {
				valid.push(i);
				continue;
			}
			if (!valid.isEmpty()) {
				int pos = valid.pop();
				dist[i] = i - pos + 1;
				if (pos > 0 && exp.charAt(pos - 1) == ')')
					dist[i] += dist[pos - 1];
			}
		}
		for (int i = 0; i < exp.length(); i++) {
			if (dist[i] > max) {
				max = dist[i];
				occ = 1;
				continue;
			}

			if (dist[i] == max)
				occ++;
		}
		if (max == 0)
			occ = 1;
		out.println(max + " " + occ);

		out.flush();
		out.close();

	}
}