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
#!usr/bin/env python3
from collections import defaultdict, deque
from heapq import heappush, heappop
from itertools import permutations, accumulate
import sys
import math
import bisect
def LI(): return [int(x) for x in sys.stdin.readline().split()]
def I(): return int(sys.stdin.readline())
def LS():return [list(x) for x in sys.stdin.readline().split()]
def S():
    res = list(sys.stdin.readline())
    if res[-1] == "\n":
        return res[:-1]
    return res
def IR(n):
    return [I() for _ in range(n)]
def LIR(n):
    return [LI() for _ in range(n)]
def SR(n):
    return [S() for _ in range(n)]
def LSR(n):
    return [LS() for _ in range(n)]

sys.setrecursionlimit(1000000)
mod = 1000000007

def solve():
    n = I()
    a = LI()
    b = LI()
    cnt = defaultdict(lambda : 0)
    for i in a:
        cnt[i] += 1
    for i in b:
        if not cnt[i]:
            continue
        cnt[i] += 1
    for i in cnt.values():
        if i > n:
            print("No")
            return
    r = n-1
    while 1:
        update = False
        for l in range(n):
            while r >= 0 and a[r] != b[r]:
                r -= 1
            if a[l] == b[l]:
                update = True
                b[l],b[r] = b[r],b[l]
                r -= 1
        if not update:
            break
    print("Yes")
    print(*b)
    return

#Solve
if __name__ == "__main__":
    solve()
