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 C++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <stdlib.h>
#include <iostream>
#include <math.h>
#include <algorithm>
#include <string>
#include <string.h>
#include <vector>
#include <functional>
#include <queue>
#include <unordered_set>
#include <climits>
#include <set>
#include <list>
#include <cmath>
#include <map>

using namespace std;
typedef long long ll;

int main()
{
	int N;
	int A[200200];
	int B[200200];
	int ACount[200200];
	int BCount[200200];
	for(int i = 0; i < 200200; i++){
		ACount[i] = 0;
		BCount[i] = 0;
	}
	cin >> N;
	for(int i = 0; i < N; i++){
		cin >> A[i];
		ACount[A[i]]++;
	}
	for(int i = 0; i < N; i++){
		cin >> B[i];
		BCount[B[i]]++;
	}
	int over = 0;
	for(int i = 1; i <= N; i++){
		if(ACount[i] > 0 && BCount[i] > 0){
			over = max(over, ACount[i] + BCount[i]);
		}
	}
	if(over > N){
		cout << "No" << endl;
		return 0;
	}else{
		cout << "Yes" << endl;
	}
	
	for(int t = 0; t < 6000; t++){
		for(int i = 0; i < N; i++){
			if(A[i] == B[i]){
				int next = i+1 == N ? 0 : i+1;
				int temp = B[next];
				B[next] = B[i];
				B[i] = temp;
			}
		}
	}
	
	for(int i = 0; i < N; i++){
		cout << B[i] << " ";
	}
	cout << endl;
	
	return 0;
}