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 = sys.stdin.readline
sys.setrecursionlimit(10 ** 7)

n = int(input())
a = list(map(int, input().split()))
b = list(map(int, input().split()))

c = [0] * (n + 1)
d = [0] * (n + 1)
for i in range(n):
    c[a[i]] += 1
    d[b[i]] += 1
# 鳩の巣原理
# どんなに頑張っても1個以上重なる
for i in range(n + 1):
    if c[i] + d[i] > n:
        print('No')
        sys.exit(0)

import itertools
c = list(itertools.accumulate(c))
d = list(itertools.accumulate(d))

max_gap = 0
for i in range(1, n + 1):
    max_gap = max(max_gap, c[i] - d[i-1])

ans = [0] * n
for i in range(n):
    ans[(i + max_gap) % n] = str(b[i])

print('Yes')
print(' '.join(ans))
