Problem:
Score : 600 points

Problem Statement

A bracket sequence is a string that is one of the following:

An empty string;

The concatenation of (, A, and ) in this order, for some bracket sequence A ;

The concatenation of A and B in this order, for some non-empty bracket sequences A and B /

Given are N strings S_i. Can a bracket sequence be formed by concatenating all the N strings in some order?

Constraints

1 \leq N \leq 10^6

The total length of the strings S_i is at most 10^6.

S_i is a non-empty string consisting of ( and ).

Input

Input is given from Standard Input in the following format:

N
S_1
:
S_N

Output

If a bracket sequence can be formed by concatenating all the N strings in some order, print Yes; otherwise, print No.

Sample Input 1

2
)
(()

Sample Output 1

Yes

Concatenating (() and ) in this order forms a bracket sequence.

Sample Input 2

2
)(
()

Sample Output 2

No

Sample Input 3

4
((()))
((((((
))))))
()()()

Sample Output 3

Yes

Sample Input 4

3
(((
)
)

Sample Output 4

No

Solution in PyPy3 (7.3.0) (judge verdict: Accepted)

N=int(input())
S=[]
L=0
R=0
for i in range(N):
    s=input()
    n=len(s)
    data=[s[j] for j in range(n)]
    l=0
    r=0
    yabai=float("inf")
    for j in range(n):
        l+=(s[j]=="(")
        r+=(s[j]==")")
        yabai=min(l-r,yabai)
    S.append([yabai,l-r])
    L+=l
    R+=r

if L!=R:
    print("No")
else:
    first=[]
    last=[]
    for i in range(N):
        yabai,gap=S[i]
        if gap>0:
            first.append(S[i])
        else:
            last.append([gap-yabai,yabai])
    first.sort(reverse=True)
    last.sort(reverse=True)
    G=0
    for i in range(len(first)):
        yabai,gap=first[i]
        if 0>G+yabai:
            print("No")
            exit()
        else:
            G+=gap
    for j in range(len(last)):
        gapminus,yabai=last[j]
        if 0>G+yabai:
            print("No")
            break
        else:
            G+=gapminus+yabai
    else:
        print("Yes")