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)

from collections import Counter, defaultdict, deque
n = int(input())
A = list(map(int, input().split()))
Borigin = list(map(int, input().split()))

Acnt = Counter(A)
NG = defaultdict(int)
for k, v in Acnt.items():
    NG[k] = n-v+1

Bcnt = Counter(Borigin)
B = deque(Borigin.copy())
newB = [None]*n
cnt = 0
is_skip = 0
for i, a in enumerate(A):
    while True:
        cnt += 1
        if 10**6 < cnt:
            is_skip = 1
            break

        b = B.pop()
        if a == b:
            B.appendleft(b)
            continue
        if Bcnt[b]:
            break
        B.append(b)
    if is_skip:
        break
    Bcnt[b] -= 1
    newB[i] = b

if not is_skip:
    print('Yes')
    print(*newB)
    exit()

B = deque(Borigin.copy())
newB = [None]*n
cnt = 0
for i, a in enumerate(A):
    while True:
        cnt += 1
        if 10**6 < cnt:
            print('No')
            exit()
        b = B.popleft()
        if a == b:
            B.append(b)
            continue
        if Bcnt[b]:
            break
        B.append(b)
    Bcnt[b] -= 1
    newB[i] = b

print('Yes')
print(*newB)