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: Wrong Answer)

N = int(input())
S = [list(input()) for i in range(N)]
check1 = 0
check2 = 0

UPDOWN = [0]*N
LOWEST = [0]*N


for i in range(N):
    updown=0
    lowest=0
    
    for j in S[i]:
        if j == ")":
            updown -= 1
        else:
            updown += 1
        lowest = min(updown,lowest)
    
    UPDOWN[i]+=updown
    LOWEST[i]+=lowest
    
temp = list(zip(UPDOWN,LOWEST))
temp2 = sorted(temp, key=lambda x:(-x[0], x[1]))

position = 0
check = 0

for i in range(N):
    if position + temp2[i][1]<0:
        check += 1
        break
    position += temp2[i][0]


if check == 0 and position == 0:
    print("Yes")
else:
    print("No")