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: Wrong Answer)

n = gets.to_i
a = gets.split.map(&:to_i)
b = gets.split.map(&:to_i)
at = a.tally
bt = b.tally
ok = true
n.times do |i|
  ac = at[i + 1] || 0
  bc = bt[i + 1] || 0
#  puts "#{ac} #{bc}"
  ok = false if ac + bc > n
end
puts ok ? "Yes" : "No"
if ok
  ans = []
  at.each do |k, v|
    x = k + 1
    x = 1 if x > n
    v.times do
      while (bt[x] || 0) == 0 do
        x += 1
        x = 1 if x > n
      end
      ans.push(x)
      bt[x] -= 1
    end
  end
  puts ans.join(" ")
end