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
def main():
    N = int(input())
    A = list(map(int, input().split()))
    B = list(map(int, input().split()))
    cnt = [0] * (N + 1)
    for n in (A + B):
        cnt[n] += 1
    if max(cnt) > N:
        print('No')
        return
    # Bを逆順にすると、真ん中あたりで重複のペアが生じうる。そこを解消する。
    B = B[::-1]
    dup, dup_indices = -1, list()
    for i in range(N):
        if A[i] == B[i]:
            dup = A[i]
            dup_indices.append(i)
    if dup == -1:
        # 同じ数字のペアがないので、そのまま出力
        print('Yes')
        print(' '.join(map(str, B)))
        return
    # 真ん中あたりの重複ペア部分を他と入れ替える
    indices = list()
    for i in range(N):
        if A[i] != dup and B[i] != dup:
            indices.append(i)
    indices = indices[:len(dup_indices)]
    for i, j in zip(dup_indices, indices):
        B[i], B[j] = B[j], B[i]
    print('Yes')
    print(' '.join(map(str, B)))


if __name__ == '__main__':
    main()