Problem:
Score : 600 points

Problem Statement

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.

Sample Input 1

6
1 1 1 2 2 3
1 1 1 2 2 3

Sample Output 1

Yes
2 2 3 1 1 1

Sample Input 2

3
1 1 2
1 1 3

Sample Output 2

No

Sample Input 3

4
1 1 2 3
1 2 3 3

Sample Output 3

Yes
3 3 1 2

Solution in PyPy3 (7.3.0) (judge verdict: Wrong Answer)

import sys,heapq
input = sys.stdin.buffer.readline

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

edge = [[] for i in range(n+1)]
for i in range(n):
    edge[a[i]].append(i)

res = [0]*n

dg = 10**6

c = list(range(n+1))
d = list(range(n+1))
for e in b:
    c[e] += dg
    d[e] += dg

c.sort(reverse = True)

nxt = 1
for i in range(n+1):
    if c[i] < dg:
        break
    cnt,val = divmod(c[i],dg)
    idx = c[nxt]%dg
    while cnt > 0:
        if len(edge[idx]) > 0:
            q = edge[idx].pop()
            res[q] = val
            cnt -= 1
        else:
            nxt += 1
            if nxt > n:
                break
            idx = c[nxt]%dg
            if len(edge[idx]) == 0:
                break


for e in res:
    if e == 0:
        continue
    d[e] -= dg



tank = []
for e in d:
    while e > dg:
        tank.append(e%dg)
        e -= dg

for i in range(n):
    if res[i] == 0:
        val = tank.pop()
        res[i] = val

flag = True
for i in range(n):
    if res[i] == a[i]:
        flag = False


if flag:
    print("Yes")
    print(*res)
else:
    print("No")