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 solve(n):
    a = list(map(int, input().split()))
    b = list(map(int, input().split()))[::-1]
    ind = [-1] * (n + 1)
    for i in range(n):
        if i == 0:
            ind[b[i]] = 0
        elif b[i] != b[i - 1]:
            ind[b[i]] = i
    cnt = 0
    for i in range(n):
        if a[i] != b[i]:
            continue
        else:
            swap = 0
            for j in range(b[i] - 1, 0, -1):
                if ind[j] != -1 or ind[j] >= ind[j - 1]:
                    swap = ind[j]
                    ind[j] += 1
                    break
            if n <= swap:
                swap = cnt
                cnt += 1
                cnt %= n
            b[i], b[swap] = b[swap], b[i]
    return a, b


def main():
    n = int(input())
    a, b = solve(n)
    if all(a[i] != b[i] for i in range(n)):
        print("Yes")
        print(" ".join(map(str, b)))
    else:
        print("No")


if __name__ == '__main__':
    main()
