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: Runtime Error)


import bisect

n = int(input())

a = [int(i) for i in input().split()]
b = [int(i) for i in input().split()]

ab = [[0 for _ in range(2)] for _ in range(10)]

for i in range(n):
    ab[a[i]][0] += 1
    ab[b[i]][1] += 1

flag = 0
for i in range(10):
    if(ab[i][0] + ab[i][0] > n):
        flag = 1

if (flag == 1):
    print("No")
else:
    print("Yes")
    c = sorted(a)
    d = sorted(b)
    kotae = []
    e = [0 for _ in range(10)]
    for i in range(n):
        kotae.append(d[bisect.bisect_left(c,a[i])+e[a[i]]])
        e[a[i]] += 1
    kotae3 =[]
    for i in range(n//2+1,n):
        kotae3.append(kotae[i])
    for i in range(n//2+1):
        kotae3.append(kotae[i])
    kotae2 = [str(a) for a in kotae3]
    kotae2 = " ".join(kotae2)
    print(kotae2)