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
n = int(input())
a = list(map(int,input().split()))
b = list(map(int,input().split()))
b.reverse()
c = []
cs = -1
e = -1
for i in range(n):
  if a[i] > b[i]:
    c.append(1)
  elif a[i] == b[i]:
    c.append(0)
    e = a[i]
    if cs == -1:
      cs = i
  else:
    c.append(-1)
#print(a)
#print(b)
#print(c)
ans = 0
db = 0
for i in range(n):
  if a[i] == e:
    ans += 1
  if b[i] == e:
    ans += 1
  if a[i] == b[i]:
    db += 1
if ans > n:
  print('No')
  exit()
print('Yes')
cnt = 0
#print(b)
#print(cs,db)
for i in range(n):
  if cnt == db:
    break
  #print(i,a[i],b[i])
  if a[i] != e and b[i] != e:
    b[i],b[cs+cnt] = b[cs+cnt],b[i]
    cnt += 1
    #print(b)
print(' '.join(map(str,b)))