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
# bを必要なだけ(strideの分)右にずらせばよい
def main():
    n = int(input())
    a = list(map(int, input().split()))
    b = list(map(int, input().split()))

    ai, bi = 0, 0
    stride = 0
    while ai < n and bi < n:
        ax, bx = a[ai], b[bi]
        if ax == bx:
            cnt = 2

            ai += 1
            while ai < n and a[ai] == ax:
                cnt += 1
                ai += 1

            bi_copy = bi
            bi += 1
            while bi < n and b[bi] == bx:
                cnt += 1
                bi += 1

            if cnt > n:
                print('No')
                exit()

            if stride < ai - bi_copy:
                stride = ai - bi_copy

        elif ax < bx:
            ai += 1
        else:
            bi += 1

    print('Yes')
    print(' '.join(map(str, b[n-stride:] + b[:n-stride])))

main()