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
#: Author - Soumya Saurav
import sys,io,os,time
from collections import defaultdict
from collections import Counter
from collections import deque
from itertools import combinations
from itertools import permutations
import bisect,math,heapq
alphabet = "abcdefghijklmnopqrstuvwxyz"

input = sys.stdin.readline

########################################
n = int(input())
a = list(map(int , input().split()))
b = list(map(int , input().split()))
b = b[::-1]
c = -1
for i in range(n):
	if a[i] == b[i]:
		c = a[i]
		break
l = 0
for i in range(n):
	if a[i] == c and b[i] == c:
		l = i
		break
r = -1
for i in range(n-1,-1,-1):
	if a[i] == c and b[i] == c:
		r = i
		break
for i in range(n):
	if a[i] != c and b[i] != c and l<=r:
		b[i] , b[l] = b[l], b[i]
		l+=1

if l <= r:
	print("No")
	exit()
print("Yes")
print(*b)











'''
# Wrap solution for more recursion depth
#submit as python3
import collections,sys,threading
sys.setrecursionlimit(10**9)
threading.stack_size(10**8)



threading.Thread(target=solve).start()
'''
