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 PYTHON3):

Solution:
import sys

OPEN, CLOSE = '(', ')'


def magic(data):
    num_of_opened, i = 0, 0
    num_of_chars = len(data)
    indices_of_opened = []
    while i < num_of_chars:
        char = data[i]
        if char == OPEN:
            indices_of_opened.append(i)
            num_of_opened += 1
        elif num_of_opened > 0 and char == CLOSE:
            num_of_opened -= 1
            yield indices_of_opened.pop()
            yield i
        i += 1


def solve(string):
    group, prev = 0, 0
    maxgroup, maxgroupcount = 0, 0
    for index in sorted(magic(tuple(string))):
        sub = index - prev
        if sub == 1 or sub == index:
            group += 1
        elif group > 0:
            if maxgroup == group: maxgroupcount += 1
            if maxgroup < group:
                maxgroup = group
                maxgroupcount = 1
            group = 1
        prev = index

    if maxgroup < group:
        maxgroup = group
        maxgroupcount = 1
    elif maxgroup == group:
        maxgroupcount += 1

    return "%s %s" % (maxgroup, maxgroupcount)


print(solve(sys.stdin.readline()))