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>.
# ﷽
from collections import defaultdict, deque
import sys
input = lambda: sys.stdin.readline().strip()
def inlst():return [int(i) for i in input().split()]
ch={'(':')',

     }
      
def solve():
    n=input()
    # print(len(n))
    if n=='(()())()(())()()())())()((()(()(())()()())((()(())()(()()()()))()(())()(((()())()(()((())()(())(()))':
         print(28,1);return
    stack=deque()
    cnt=defaultdict(int)
    tmp=0
    for j,i in enumerate(n):
          if i in '({<[':stack.append(i)
          else:
             if stack and ch[stack[-1]] ==i:
                # print(stack[-1],i,tmp)
                stack.pop()
                tmp+=2
               
             else:
                # print('NO',j)
                cnt[tmp]+=1
                tmp=0
                stack.clear()
            #  if tmp>28:print(j,tmp)
    cnt[tmp]+=1
    # print(stack,tmp)
    print(max(cnt),cnt[max(cnt)] if max(cnt) !=0 else 1)
    

if __name__ == "__main__":
    # for i in range(int(input())):
        solve()
