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
input = lambda: sys.stdin.readline().rstrip("\r\n")

n = int(input())
a = [int(x) for x in input().split()]
b = [int(x) for x in input().split()][::-1]

ng = []
for i in range(n):
    if a[i] == b[i]:
        ng.append(i)

if len(ng) == 0:
    print('Yes')
    print(*b)
    sys.exit()

min_ng = min(ng)
max_ng = max(ng)

left = 0
right = n - 1

for i in ng:
    if b[i] != b[left] and a[left] != b[i]:
        b[i], b[left] = b[left], b[i]
        left += 1
        continue
    if b[i] != b[right] and a[right] != b[i]:
        b[i], b[right] = b[right], b[i]
        right -= 1

flag = 1

for i in range(n):
    if a[i] == b[i]:
        flag = 0

if flag:
    print('Yes')
    print(*b)
else:
    print('No') 

