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:
RI = lambda : [int(x) for x in input().split()]
rw = lambda : [input().strip()]


s=list(input().strip())

dp=[0]*(len(s))

def recursor(n):
    if(n==0):
        return(0)
    elif(n==1):
        if(s[n-1]=='('):
            return(2)
        else:
            return(0)
    elif(n>=2):
        if(s[n-1]=='('):
            return(dp[n-2]+2)
        else:
            if(n-dp[n-1]-1>=0 and s[n-dp[n-1]-1]=='('):
                if(n-dp[n-1]-2>=0):
                    return(dp[n-1]+dp[n-dp[n-1]-2]+2)
                else:
                    return(dp[n-1]+2)
            else:
                return(0)

maxo=0
c=0
for i in range(len(s)):
    if(s[i]=='('):
        continue
    dp[i]=recursor(i)
    length_temp=dp[i]
    if(length_temp>maxo):
        maxo=length_temp
        c=1
    elif(length_temp==maxo and maxo!=0):
       c+=1
                        #print(dp)

if(c!=0):
    print(maxo ,end=" ")
    print(c)
else:
    print("0 1")