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
from collections import Counter,defaultdict,deque
from heapq import heappop,heappush
from bisect import bisect_left,bisect_right 
import sys,math,itertools,fractions,pprint
sys.setrecursionlimit(10**8)
mod = 10**9+7
INF = float('inf')
def inp(): return int(sys.stdin.readline())
def inpl(): return list(map(int, sys.stdin.readline().split()))

import random
n = inp()
a = inpl()
b = inpl()
c = [0] * (n+1)
for i in range(n):
    c[a[i]] += 1
    c[b[i]] += 1
if max(c) > n:
    print('No')
    quit()
q = deque([])
for i in range(n):
    if a[i] == b[i]:
        q.append(i)
while q:
    now = q.popleft()
    nx = random.randint(0,n-1)
    if a[now] != b[nx] and a[nx] != b[now]:
        b[now],b[nx] = b[nx],b[now]
    else: q.append(now)
print('Yes')
print(*b)