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: Runtime Error)

n = gets.chomp.to_i
a = gets.chomp.split(' ').map(&:to_i)
b = gets.chomp.split(' ').map(&:to_i)
ca = a.tally
cb = b.tally

if (1..n).any? { |i| ca.fetch(i, 0) + cb.fetch(i, 0) > n }
  puts 'No'
else
  puts 'Yes'

  ans = Array.new(n)

  (1..n).each do |i|
    next unless ca[i] && cb[i]

    idx = a.bsearch_index { |x| x >= i }

    k = 1
    (0...ca[i]).each do |j|
      loop do
        if k == i || !cb[k]
          k += 1
          k = 1 if k > n
        end
        break if ca[k]
      end

      ans[idx + j] = k
      cb[k] -= 1
      cb.delete(k) if cb[k] == 0
    end
  end

  (0...n).each do |j|
    next if ans[j]
    k = cb.keys[0]
    ans[j] = k
    cb[k] -= 1
    cb.delete(k) if cb[k] == 0
  end

  puts ans.join(' ')
end