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
N = int(input())
A = list(map(int,input().split()))
B = list(map(int,input().split()))
RB = B[::-1]

start = v = -1
seq = 0
for i,(a,b) in enumerate(zip(A,RB)):
    if a==b:
        v = a
        seq += 1
        if start == -1:
            start = i

if v == -1:
    print('Yes')
    print(*RB)
    exit()

if A.count(v) + B.count(v) > N:
    print('No')
    exit()

from bisect import bisect_left,bisect
al = bisect_left(A,v)
br = bisect(B,v)
rbl = N - br

m = min(seq,start,al,rbl)
RB[:m],RB[start:start+m] = RB[start:start+m],RB[:m]
seq -= m
start += m
if seq:
    RB[start:start+seq],RB[-seq:] = RB[-seq:],RB[start:start+seq]

assert all(a!=b for a,b in zip(A,RB))
print('Yes')
print(*RB)