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(int,readline().split())
*b, = map(int,readline().split())
b = b[::-1]

same = [i for i in range(n) if a[i] == b[i]]
if same:
    # 以下すべて半開区間
    v = a[same[0]] # 重なる値は一種類しかない。これを v とおく。
    r = same[0]
    s = same[-1]+1
    vidx = [i for i in range(n) if a[i]==v or b[i]==v]
    p = vidx[0]
    q = vidx[-1]+1
    # [p,q) でかぶる。[r,s)でどちらかが値 v をとる
    if q > n-s+r+p: # かぶりを吹っ飛ばせない
        print("No")
        exit()
    # かぶった部分を左、右によける
    if p >= s-r:
        b[r:s],b[:s-r] = b[:s-r],b[r:s]
    else:
        b[r:r+p],b[:p] = b[:p],b[r:r+p]
        b[r+p:s],b[n-s+r+p:] = b[n-s+r+p:], b[r+p:s]

print("Yes")
print(*b)

