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()))

T = [(A[i], i) for i in range(N)]+[(B[i], N+i) for i in range(N)]
T.sort()

P = [-1]*(2*N)
for i in range(N):
  if T[i][0]==T[i+N][0]:
    print("No")
    exit(0)
  P[T[i][1]] = T[i+N][1]
  P[T[i+N][1]] = T[i][1]

bp1 = N
for ap1 in range(N):
  if P[ap1]//N==1:
    continue
  while P[bp1]//N==0:
    bp1 += 1
  ap2 = P[ap1]
  bp2 = P[bp1]
  if A[ap1]!=B[bp1-N] and A[ap2]!=B[bp2-N]:
    P[ap1] = bp1
    P[bp1] = ap1
    P[ap2] = bp2
    P[bp2] = ap2
  else:
    P[ap1] = bp2
    P[bp1] = ap2
    P[ap2] = bp1
    P[bp2] = ap1

print("Yes")
print(" ".join(str(B[P[i]-N]) for i in range(N)))
