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
mod=10**9+7 ; inf=float("inf")
from math import sqrt, ceil
from collections import deque, Counter, defaultdict #すべてのkeyが用意されてる defaultdict(int)で初期化
input=lambda: sys.stdin.readline().strip()
sys.setrecursionlimit(11451419)
from decimal import ROUND_HALF_UP,Decimal as D  #変換後の末尾桁を0や0.01で指定
  #Decimal((str(0.5)).quantize(Decimal('0'), rounding=ROUND_HALF_UP))
from functools import lru_cache
from bisect import bisect_left as bileft, bisect_right as biright
from fractions import Fraction as F  #F(a,b)で正確なa/b
#メモ化再帰defの冒頭に毎回 @lru_cache(maxsize=10**10)
#引数にlistはだめ
#######ここまでテンプレ#######
#ソート、"a"+"b"、再帰ならPython3の方がいい
#######ここから天ぷら########

n=int(input())
A=list(map(int,input().split()))
B=list(map(int,input().split()))
j=0 ;bef=-1
for i in range(n):
    if bef!=A[i]:j=0
    if A[i]==B[i]:
        
        while j<n:
            if A[i]!=B[j] and A[i]!=A[j]:
                B[i],B[j]= B[j],B[i]
                break
            j+=1
        else:
            print("No");exit()
    bef=A[i]

print("Yes")
print(*B)
