Problem: 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

Solution (in PYTHON3):

Solution:
import sys
from itertools import accumulate

read = sys.stdin.read
readline = sys.stdin.readline
readlines = sys.stdin.readlines
sys.setrecursionlimit(10 ** 9)
INF = 1 << 60
MOD = 1000000007


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

    C = [0] * (N + 1)
    D = [0] * (N + 1)

    for i in range(N):
        C[A[i]] += 1
        D[B[i]] += 1

    for i in range(1, N + 1):
        if C[i] + D[i] > N:
            print('No')
            return

    print('Yes')

    C = list(accumulate(C))
    D = list(accumulate(D))

    B.reverse()

    i = 0
    while i < N and A[i] != B[i]:
        i += 1

    if i == N:
        print(*B)
        return

    n = A[i]
    leftA = C[n - 1]
    rightA = C[n]
    leftB = N - D[n]
    rightB = N - D[n - 1]

    i1 = min(leftA, leftB)
    i2 = max(leftA, leftB)
    i3 = min(rightA, rightB)
    i4 = max(rightA, rightB)

    c = 0
    for i in range(i1):
        if c == i3 - i2:
            break
        B[i], B[i2 + c] = B[i2 + c], B[i]
        c += 1

    for i in range(i4, N):
        if c == i3 - i2:
            break
        B[i], B[i2 + c] = B[i2 + c], B[i]
        c += 1

    print(*B)
    return


if __name__ == '__main__':
    main()