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
# Rotationによる解。
from collections import Counter

N = int(input())
As = list(map(int, input().split()))
Bs = list(map(int, input().split()))

elem_max_count = max((Counter(As) + Counter(Bs)).values())
if elem_max_count > N:
    print("No")
    exit()

# Bs[-index:] > As[:index] を満たすようなindexを二分探索
# 参考: https://twitter.com/catupper/status/1305163805802917889?s=20

ok = 0
ng = N + 1
while ng - ok > 1:
    mid = (ok + ng) // 2
    isok = True
    for b, a in zip(Bs[-mid:], As[:mid]):
        if b <= a:
            isok = False
            break
    
    if isok:
        ok = mid
    else:
        ng = mid

print("Yes")
print(*(Bs[-ok:] + Bs[:-ok]))