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
# ()の問題は折れ線で捉えると良い

N = int(input())
ls = [] # 増減が正のもの(0を含む)
rs = [] # 増減が負のもの

tot = 0
for i in range(N):
    s = input()
    h = 0
    b = 0
    for c in s:
        if c == "(":
            h += 1
        else:
            h -= 1
        b = min(b, h)
    if h >= 0:
        # 折れ線に対して上っていく奴ら
        ls.append([b, h])
    else:
        # 折れ線に対して下っていく奴ら
        # 右から(hから見るので、最下点は-hする)
        rs.append([b-h, -h])
    tot += h

ls.sort(key = lambda x:x[0], reverse=True)
rs.sort(key = lambda x:x[0], reverse=True)

def check(s):
    h = 0
    for p in s:
        if h+p[0] < 0:
            return False
        h += p[1]
    return True

if check(ls) and check(rs) and tot == 0:
    print("Yes")
else:
    print("No")