Problem: 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.

Examples

Input

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


Output

Yes
2 2 3 1 1 1


Input

3
1 1 2
1 1 3


Output

No


Input

4
1 1 2 3
1 2 3 3


Output

Yes
3 3 1 2

Solution (in PYTHON3):

Solution:
import sys
input = sys.stdin.readline
 
t = int(input())

a = list(map(int,input().split()))
b = list(map(int,input().split()))

b = b[::-1]

el = -(10**99)
ok = 1

pref, aft = -1,9999999999
pos = set()

for i in range(t):
  if a[i]==b[i]:
    pos.add(i)
    el = a[i]
    ok = 0

for i in range(t):
  if a[i]==el or b[i]==el:
    break
  pref = i
  
for i in range(t):
  if a[i]>el and b[i]!=el:
    aft = i
    break

if ok==1:
  print('Yes')
  print(*b)
  exit()
  
else:
  if len(pos)> 1+pref +(t-aft):
    print('No')
    
  else:
    try:
      for i in range(pref+1):
        x = pos.pop()
        b[i], b[x] = b[x], b[i]

      for i in range(aft,t):
        x = pos.pop()
        b[i], b[x] = b[x], b[i]
    except:
      pass
      
    print('Yes')
    print(*b)