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)

require 'set'; require 'prime'; require 'pp'
INF=Float::INFINITY; MOD=10**9+7
n = gets.chomp.to_i
a = gets.chomp.split.map(&:to_i)
b = gets.chomp.split.map(&:to_i)

left = []
right = []

i = 0
j = 0
si = 0
sj = 0
while i < n && j < n
  while a[i] == a[i+1]
    i += 1
  end
  while b[j] == b[j+1]
    j += 1
  end
  if a[i] > b[j]
    j += 1
    sj = j
  elsif a[i] < b[j]
    i += 1
    si = i
  else
    left.push j - si + 1
    right.push i - sj + 1

    j += 1
    sj = j
    i += 1
    si = i
  end
end

# p left
# p right
lmax = left.max
rmax = right.max
if left.empty?
  puts b.join(' ')
elsif lmax + rmax > n
  puts 'No'
else
  puts 'Yes'
  rmax.times do
   b.unshift b.pop
  end
  puts b.join(' ')
end