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
import sys
import heapq
from functools import lru_cache


# \n
def input():
    return sys.stdin.readline().rstrip()


def main():
    N = int(input())
    A = list(map(int, input().split()))
    B = list(map(int, input().split()))

    n = N // 3

    @lru_cache(maxsize=None)
    def diff(s):
        count = 0
        for i in range(N):
            if A[i] == B[(i + s) % N]:
                count += 1

        if count == 0:
            print("Yes")
            L = []
            for i in range(N):
                L.append(B[(i + s) %N])
            print(*L)
            exit()

        return count

    l = 0
    r = N - 1
    diff(l)
    diff(r)

    while r - l > 2:
        a = diff(l + (r - l) // 3)
        b = diff(l + 2 * (r - l) // 3)
        if a > b:
            l = l + (r - l) // 3
        else:
            r =l + 2 * (r - l) // 3
    diff((l+r)//2)

    print("No")


if __name__ == "__main__":
    main()
