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
# region header
import sys, bisect, math, itertools, heapq, collections
from operator import itemgetter
# a.sort(key=itemgetter(i)) # i番目要素でsort
from functools import lru_cache
import copy
# @lru_cache(maxsize=None)
# sys.setrecursionlimit(10**8)
input = sys.stdin.readline
INF = float('inf')
mod = 10**9 + 7
eps = 10**-7
# endregion
# region input function


def inp():
    '''
    一つの整数
    '''
    return int(input())


def inpl():
    '''
    一行に複数の整数
    '''
    return list(map(int, input().split()))


def str_inp():
    '''
    文字列をリストとして読み込む
    '''
    return list(input()[:-1])


# endregion

n = inp()
a = inpl()
b = inpl()
a_digest = collections.defaultdict(int)
b_digest = collections.defaultdict(int)
for i in range(n):
    a_digest[a[i]] += 1
    b_digest[b[i]] += 1
for i in range(1, n + 1):
    if a_digest[i] > n - b_digest[i]:
        print("No")
        exit()
print("Yes")
b.sort(reverse=True)
ptr = 0
for i in range(n):
    if a[i] == b[i]:
        while a[i] == b[ptr] or a[ptr] == b[i]:
            ptr += 1
            if ptr == n:
                ptr = 0
        b[ptr], b[i] = b[i], b[ptr]
print(*b)
