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 <iostream>
#include <string>
#include <sstream>
#include <iomanip> 
#include <math.h>
#include <stdio.h>
#include <string.h>
#include <queue>
#include <stack>
#include <vector>
#include <map>
#include <set>
#include <functional>
#include <algorithm>
#include <unordered_map>
#include <unordered_set>
#include <bitset>
#include <complex>

using namespace std;

//#pragma GCC optimize("Ofast")
//#pragma GCC optimization("unroll-loops, no-stack-protector")
//#pragma GCC target("avx,avx2,fma")


int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);

	int n, i, j, k;
	cin >> n;
	vector<int> a(n), b(n), c;
	map<int, int> ca, cb;
	for (i = 0; i < n; i++) {
		cin >> a[i];
		ca[a[i]]++;
	}
	for (i = 0; i < n; i++) {
		cin >> b[i];
		cb[b[i]]++;
	}

	bool ans;

	ans = true;
	for (auto p : ca) {
		if (p.second * 2 > n) ans = false;
	}

	if (ans) {
		cout << "Yes\n";
		queue<int> q;
		for (auto p : b) q.push(p);

		for (auto p : a) {
			while (q.front() <= p) {
				q.push(q.front());
				q.pop();
			}

			c.push_back(q.front());
			q.pop();
		}
		for (auto p : c) cout << p << " ";
		cout << "\n";
	}

	if (!ans) cout << "No\n";
		

	return 0;
}