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:
# @author: guoyc
# @date: 2018/7/20

s = input()

N = len(s)

# dp = [0] * (N + 1)
#
# ml = 0
# cnt = 1
#
# for i in range(1, N):
#     if s[i] == '(':
#         continue
#     if s[i - 1] == '(':
#         dp[i] = 2 + dp[i - 2]
#     elif (i - 1) - dp[i - 1] >= 0:
#         if s[(i - 1) - dp[i - 1]] == '(':
#             dp[i] = dp[i - 1] + 2 + dp[(i - 1) - dp[i - 1] - 1]
#     if dp[i] > ml:
#         ml = dp[i]
#         cnt = 1
#     elif dp[i] == ml:
#         cnt += 1
#
# if ml == 0:
#     print(0, 1)
# else:
#     print(ml, cnt)

st = [-1]

ml, cnt = 0, 1

for i, c in enumerate(s):
    if c == '(':
        st.append(i)
    else:
        if len(st) > 1:
            st.pop(-1)
            ll = i - st[-1]
            if ll > ml:
                ml, cnt = ll, 1
            elif ll == ml:
                cnt += 1
        else:
            st[0] = i

print(ml, cnt)