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())
left = []
midplus = []
midzero = []
midminus = []
right = []
L = []
R = []

for i in range(n):
    s = input()
    l = 0
    r = 0
    for x in s:
        if x == '(':
            l += 1
        else:
            if l > 0:
                l -= 1
            else:
                r += 1
    if l > 0 and r == 0:
        left.append((l, r))
    elif l > 0 and r > 0:
        if l > r:
            midplus.append((r, l))  # a,b-a
        elif l == r:
            midzero.append((r,l))
        else:
            midminus.append((r, l))
    elif l == 0 and r > 0:
        right.append((l, r))
    L.append(l)
    R.append(r)

if sum(L) != sum(R):
    print('No')
    exit()

A = 0
B = 0
for x in left:
    A += x[0]
for x in right:
    B += x[1]

midplus = sorted(midplus, key=lambda x: (x[0], -x[1]))
midminus = sorted(midminus, key=lambda x: -x[1])
mid = midplus + midzero + midminus

l = A
r = 0
for a, b in mid:
    if l < a:
        print('No')
        exit()
    l -= a
    l += b

print('Yes')