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.StringTokenizer;


public class Main {
    static char[] input;
    static int[] states;

    /**
     * @param args
     * @throws IOException 
     */
    public static void main(String[] args) throws IOException {
        // TODO Auto-generated method stub
        BufferedReader in=new BufferedReader(new InputStreamReader(System.in));
        PrintWriter out=new PrintWriter(System.out);
        char[] input = in.readLine().toCharArray();
        int[] x = new int[input.length];
        int num = 0, first = 0;
        int maxLen = 0, maxCount = 1;
        for (int i = 0; i < input.length; i++) {
            if (input[i] == '(') {
                x[num] = first;
                num++;
                first = i + 1;
            } else {
                if (num == 0) {
                    first = i + 1;
                    continue;
                }
                int x1 = x[num - 1], x2 = i;
                if (x2 - x1 + 1 > maxLen) {
                    maxLen = x2 - x1 + 1;
                    maxCount = 1;
                } else if (x2 - x1 + 1 == maxLen) {
                    maxCount++;
                }
                first = x1;
                num--;
            }
        }
        out.print(maxLen + " " + maxCount);
        out.close();
    }

}