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 bisect

n=int(input())
opens=0
closes=0
ls=[]
rs=[]
for _ in range(n):
  s=input()
  cnt1=0
  cnt2=0
  for i in range(len(s)):
    if s[i]=='(':
      cnt1+=1
    else:
      if cnt1==0:
        cnt2+=1
      else:
        cnt1-=1
  if cnt2==0:
    opens+=cnt1
  elif cnt1==0:
    closes+=cnt2
  else:
    ls.append(cnt2)
    rs.append(cnt1)
ls=sorted(ls)
rs=sorted(rs)
lefts=sum(ls)
rights=sum(rs)
if opens-closes!=lefts-rights:
  print('No')
else:
  flag=False
  if len(ls)==0:
    flag=True
  for a in ls:
    if opens<a or closes<a+(rights-lefts):
      continue
    pos=bisect.bisect_right(rs,a+(rights-lefts))
    if pos!=0:
      flag=True
      break
  for d in rs:
    if opens<d+(lefts-rights) or closes<d:
      continue
    pos=bisect.bisect_right(ls,d+(lefts-rights))
    if pos!=0:
      flag=True
      break
  if flag==True:
    print('Yes')
  else:
    print('No')