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
from collections import Counter
N = int(input())
A = list(map(int,input().split()))
B = list(map(int,input().split()))

A_cnt = Counter(A)
B_cnt = Counter(B)
for i in range(1,N+1):
    if N < A_cnt[i]+B_cnt[i]:
        print("No")
        break
else:
    print("Yes")
    cnt = 0
    indexes = {}
    alreadychecked = set()
    for i in range(N):
        if A[i] not in alreadychecked:
            indexes[A[i]] = i
            alreadychecked.add(A[i])
    zure = 0
    alreadychecked = set()
    for i in range(N-1,-1,-1):
        if B[i] not in alreadychecked:
            if B[i] in indexes and indexes[B[i]] <= i:
                zure = max(zure,i-indexes[B[i]]+1)
                alreadychecked.add(B[i])

    print(*B[zure:],*B[:zure])
