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:
z,zz=input,lambda:list(map(int,z().split()))
zzz=lambda:[int(i) for i in stdin.readline().split()]
szz,graph,mod,szzz=lambda:sorted(zz()),{},10**9+7,lambda:sorted(zzz())
from string import *
from collections import *
from queue import *
from sys import *
from collections import *
from math import *
from heapq import *
from itertools import *
from bisect import *
from collections import Counter as cc
from math import factorial as f
from bisect import bisect as bs
from bisect import bisect_left as bsl
from itertools import accumulate as ac
def lcd(xnum1,xnum2):return (xnum1*xnum2//gcd(xnum1,xnum2))
def prime(x):
    p=ceil(x**.5)+1
    for i in range(2,p):
        if (x%i==0 and x!=2) or x==0:return 0
        
    return 1
def dfs(u,visit,graph):
    visit[u]=True
    for i in graph[u]:
        if not visit[i]:
            dfs(i,visit,graph)

###########################---Test-Case---#################################
"""



"""
###########################---START-CODING---##############################

st=[-1]
mx,cnt=0,0
for i,j in enumerate(z()):
    if j=='(':
        st.append(i)
    else:
        if len(st)>1:
            st.pop()
            temp=i-st[-1]
            if temp>mx:
                mx=temp
                cnt=1
            elif temp==mx:
                cnt+=1
        else:
            st[0]=i
if mx:
    print(mx,cnt)
else:
    print(0,1)