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.to_i
a = gets.split.map &:to_i
b = gets.split.map &:to_i

c = a.tally
d = b.tally

e, f = c.max_by { |k, v| v }
g, h = d.max_by { |k, v| v }

if d[e] + f > n || c[g] + h > n
  puts "No"
else
  puts "Yes"
  i = 0
  while b.length > 1
    tmp = b.pop
    if tmp != a[i]
      print "#{tmp} "
      i += 1
    else
      b.unshift tmp
    end
  end
  puts b.pop
end