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 Python (3.8.2) (judge verdict: Accepted)

import bisect, heapq
N = int(input())
A = list(map(int,input().split()))
B = list(map(int,input().split()))

for i in range(1,N+1):
    a = bisect.bisect_right(A,i) - bisect.bisect_left(A,i)
    b = bisect.bisect_right(B,i) - bisect.bisect_left(B,i)

    if a + b > N:
        print("No")
        exit()

B = B[::-1]

j, bef = 0, -1
for i in range(N):
    if A[i] != bef:
        j = 0
    if A[i] == B[i]:
        while j < N:
            if A[j] != B[i] and B[j] != A[i]:
                B[i], B[j] = B[j], B[i]
                break
            j += 1
    bef = A[i]

print("Yes")
for b in B:
    print(b,end=" ")
print("")