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.
The first line of the input file contains a non-empty string, consisting of «(» and «)» characters. Its length does not exceed 10<sup class="upper-index">6</sup>.
s = input()
stack = []
dp = [[-1, -1] for _ in range(len(s))]
# dp[i][0] = earliest index of s[i] such that [dp[i][0], i] is a valid substring
# dp[i][1] = corresponding open bracket

for i in range(len(s)):
    if s[i] == "(":
        stack.append(i)
    elif s[i] == ")":
        if len(stack) == 0:
            continue
        j = stack.pop()
        dp[i][1] = j
        if s[j-1] == ")" and dp[j-1][0] != -1:
            dp[i][0] = dp[j-1][0]
        else:
            dp[i][0] = j
max_len = 0
for i in range(len(s)):
    if dp[i][0] == -1:
        continue
    if i - dp[i][0] + 1 > max_len:
        max_len = i - dp[i][0] + 1
num_substrings = 0
for i in range(len(s)):
    if dp[i][0] == -1:
        continue
    if i - dp[i][0] + 1 == max_len:
        num_substrings += 1
if max_len == 0:
    num_substrings = 1
print(max_len, num_substrings)