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
import sys
n = int(input())
si = []
for _ in range(n):
    s = input()
    l = 0
    r = 0
    count = 0
    for i in range(len(s)):
        if s[i] == "(":
            count += 1
        else:
            count = max(0, count-1)
    l = count
    count = 0
    for i in range(len(s)-1, -1, -1):
        if s[i] == ")":
            count += 1
        else:
            count = max(0, count-1)
    r = count
    si.append((l, r))

minus = []
plus = []
while si:
    a, b = si.pop()
    if a <= b:
        plus.append((a, b))
    else:
        minus.append((b, a))

plus.sort()
minus.sort()
countl = 0
for a, b in plus:
    if countl >= a:
        countl += b-a
    else:
        print("No")
        sys.exit()

countr = 0
for a, b in minus:
    if countr >= a:
        countr += b-a
    else:
        print("No")
        sys.exit()


if countr == countl:
    print("Yes")
else:
    print("No")
