Problem: Given are two sequences A and B, both of length N. A and B are each sorted in the ascending order. Check if it is possible to reorder the terms of B so that for each i (1 \leq i \leq N) A_i \neq B_i holds, and if it is possible, output any of the reorderings that achieve it.

Constraints

* 1\leq N \leq 2 \times 10^5
* 1\leq A_i,B_i \leq N
* A and B are each sorted in the ascending order.
* All values in input are integers.

Input

Input is given from Standard Input in the following format:


N
A_1 A_2 \cdots A_N
B_1 B_2 \cdots B_N


Output

If there exist no reorderings that satisfy the condition, print `No`.

If there exists a reordering that satisfies the condition, print `Yes` on the first line. After that, print a reordering of B on the second line, separating terms with a whitespace.

If there are multiple reorderings that satisfy the condition, you can print any of them.

Examples

Input

6
1 1 1 2 2 3
1 1 1 2 2 3


Output

Yes
2 2 3 1 1 1


Input

3
1 1 2
1 1 3


Output

No


Input

4
1 1 2 3
1 2 3 3


Output

Yes
3 3 1 2

Solution (in PYTHON3):

Solution:
n=int(input())
a=list(map(int,input().split()))
b=list(map(int,input().split()))
d={}
for i in a:
  if i in d:
    d[i][0]+=1
  else:
    d[i]=[1,0]
for i in b:
  if i in d:
    d[i][0]+=1
    d[i][1]+=1
  else:
    d[i]=[1,1]
l=[]
for i in d:
  l.append([d[i][0],i,d[i][1]])
l.sort()
l.reverse()


if l[0][0]>n:
  print("No")
else:
  print("Yes")
  l.append(l[0])
  cc=0
  ansa=[]
  i=1
  while l[i][2]==0:
    i+=1
  j=i+1
  while j!=len(l) and l[j][2]==0:
    j+=1
  ic,jc=0,0
  for t in a:
    if t==l[0][1]:
      continue
    cc+=1
    if cc>n-l[0][0]:
      ansa.append(l[0][1])
    elif l[i][1]==t:
      ansa.append(l[j][1])
      jc+=1
      if jc==l[j][2]:
        j+=1
        while j!=len(l) and l[j][2]==0:
          j+=1
        jc=0
    else:
      ansa.append(l[i][1])
      ic+=1
      if ic==l[i][2]:
        i=j
        j+=1
        while j!=len(l) and l[j][2]==0:
          j+=1
        ic=jc
        jc=0
  ans=[]
  ii=0
  for t in a:
    if t==l[0][1]:
      ans.append(l[i][1])
      ic+=1
      if ic==l[i][2]:
        i=j
        j+=1
        while j!=len(l) and l[j][2]==0:
          j+=1
        ic=jc
        jc=0
    else:
      ans.append(ansa[ii])
      ii+=1
  print(" ".join(map(str,ans)))