Problem:
Score : 600 points

Problem Statement

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.

Sample Input 1

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

Sample Output 1

Yes
2 2 3 1 1 1

Sample Input 2

3
1 1 2
1 1 3

Sample Output 2

No

Sample Input 3

4
1 1 2 3
1 2 3 3

Sample Output 3

Yes
3 3 1 2

Solution in Ruby (2.7.1) (judge verdict: Time Limit Exceeded)

n=gets.to_i
l=[0]*200001
a=gets.split.map(&:to_i)
b=gets.split.map(&:to_i)
a.each do |c|
  l[c]+=1
end
b.each do |c|
  l[c]+=1
end
lm=l.max
if lm>n
  puts "No"
  exit
else
  puts "Yes"
end
li=l.index(lm)
if !a.include?(li)||a.reverse.index(li)==0
  k=0
else
  k=n-a.reverse.index(li)
end
i,j=0,0
s=[]
jz=n
while i<n
  if b[j]&&a[(i+k)%n]!=b[j]
    s << b[j]
    b[j]=nil
    i+=1
  end
  j+=1
  if j==jz
    j=0
    b.delete(nil)
  end
end
if k==0
  bb=s
else
  bb=s[(n-k)..(n-1)]+s[0..(n-k)-1]
end
puts bb.join(" ")