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
# coding: utf-8
# Your code here!
import sys
readline = sys.stdin.readline
read = sys.stdin.read

n, = map(int,readline().split())
*a, = map(lambda x:2*int(x),readline().split())
*b, = map(lambda x:2*int(x)+1,readline().split())

c = sorted(a+b)
res = []
aa = []
bb = []
for x,y in zip(c[:n],c[n:]):
    if x%2 == 0:
        if y%2 == 0:
            aa.append((x//2,y//2))
        else:
            res.append((x//2,y//2))
    else:
        if y%2 == 0:
            res.append((y//2,x//2))
        else:
            bb.append((x//2,y//2))

while aa:
    p,q = aa.pop()
    r,s = bb.pop()
    if p != r and q != s:
        res.append((p,r))
        res.append((q,s))
    else:
        res.append((p,s))
        res.append((q,r))

from operator import itemgetter
res.sort(key = itemgetter(0))

for x,y in res:
    if x==y:
        print("No")
        exit()
print("Yes")
ans = [y for x,y in res]
print(*ans)


