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 <bits/stdc++.h>
using namespace std;
int main() {
	ios::sync_with_stdio(false);
	cin.tie(nullptr);
	int n;
	cin >> n;
	vector<int> a(n), b(n);
	for (int i = 0; i < n; i++)
		cin >> a[i];
	for (int i = 0; i < n; i++)
		cin >> b[i];
	reverse(b.begin(), b.end());
	int c = -1;
	for (int i = 0; i < n; i++) {
		if (a[i] == b[i]) {
			c = a[i];
			break;
		}
	}
	int l = find(b.begin(), b.end(), c) - b.begin();
	int r = find(b.rbegin(), b.rend(), c) - b.rbegin();
	int j = l;
	for (int i = 0; i < n; i++) {
		if ((i < l or i > r) and j <= r) {
			swap(b[i], b[j++]);
		}
	}
	if (j <= r) {
		cout << "No" << "\n";
		return 0;
	}
	cout << "Yes" << "\n";
	for (int i = 0; i < n; i++) {
		cout << b[i] << " \n"[i == n - 1];
	}
	return 0;
}