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.*;
import java.util.*;

@SuppressWarnings("unchecked")
public class Main {
    private static final long MOD = (int)1e9 + 7;

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        String s = in.next();
        int n = s.length();
        Stack<Integer> stack = new Stack<>();
        int ans = 0;
        int[] dp = new int[n];
        for (int i = 0; i < n; ++i) {
            char ch = s.charAt(i);
            if (ch == '(') {
                stack.push(i);
            }
            else if (!stack.isEmpty() && ch == ')') {
                int idx = stack.pop();
                dp[i] = i - idx + 1;
                if (idx != 0) dp[i] += dp[idx - 1];
                ans = Math.max(ans, dp[i]);
            }
        }
        System.out.print(ans + " ");
        if (ans == 0) {
            System.out.println(1);
            return;
        }
        int count = 0;
        for (int i = 0; i < n; ++i) {
            if (dp[i] == ans) ++count;
        }
        System.out.println(count);
    }
    /*
        dp[i] = length of longest good sequence ending at i;
        if (ch at i == '(') dp[i] = 0;
        else 
    */
}