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
input=sys.stdin.readline
n=int(input())
cnt=[]
for i in range(n):
	s=input().rstrip()
	l=s.count("(")
	r=s.count(")")
	minus=0
	now=0
	for j in range(len(s)):
		if s[j]=="(":
			now+=1
		else:
			now-=1
			minus=min(minus,now)
	cnt.append((l,r,s,minus))
cnt.sort(key=lambda x:x[1]-x[0])
plus=[]
minus=[]
#x[3]は最小値　x[0]-x[1]が合計値
for x in cnt:
	if x[0]-x[1]>0:
		plus.append(x)
	else:
		minus.append(x)
plus.sort(key=lambda x:-x[3])#最小値の少ない(0に近い方から)
minus.sort(key=lambda x:-(x[0]-x[1]-x[3]))
lr=0#l-r
for x in plus+minus:
	if lr+x[3]<0:
		print("No")
		exit()
	lr+=x[0]-x[1]
if lr==0:
	print("Yes")
else:
	print("No")
