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

1. An empty string;
2. The concatenation of `(`, A, and `)` in this order, for some bracket sequence A ;
3. 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`.

Examples

Input

2
)
(()


Output

Yes


Input

2
)(
()


Output

No


Input

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


Output

Yes


Input

3
(((
)
)


Output

No
from itertools import accumulate
from bisect import bisect_left
import sys
readlines = sys.stdin.buffer.readlines
readline = sys.stdin.buffer.readline

N = int(readline())

memo = [None]*N

for i,s in enumerate(readlines()):
    s = s.rstrip().decode()
    memo[i] = (min(0,min(accumulate(1 if c =='(' else -1 for c in s))),sum(1 if c =='(' else -1 for c in s))



positives = sorted((m,d) for m,d in memo if d >= 0)
negatives = sorted((m-d,-d) for m,d in memo if d < 0)

def solve():
    left = 0
    for m,d in reversed(positives):
        if left+m < 0:
            return False
        left += d
    right = 0
    for m,d in reversed(negatives):
        if right+m < 0:
            return False
        right += d
    return left == right

print('Yes' if solve() else 'No')

