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>
#define pb push_back
#define pp pair<int,int>
#define ppp pair<pp,int>
#define fi first
#define si second
#define N int(2e5+10)
using namespace std;
int cnt1[N], cnt2[N];
int main() {
#ifndef ONLINE_JUDGE
	freopen("input.txt", "r", stdin);
	freopen("output.txt", "w", stdout);
#endif

	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	cout.tie(NULL);
	cout.precision(10);
	int n;
	cin >> n;
	int a[n], b[n];
	set<ppp>s;
	vector<int>v;
	for (int i = 0; i < n; i++) {
		cin >> a[i];
		cnt1[a[i]]++;
	}
	for (int i = 0; i < n; i++) {
		cin >> b[i];
		cnt2[b[i]]++;
	}
	for (int i = 0; i < n; i++) {
		s.insert({{n - cnt1[b[i]], n - cnt2[b[i]]}, b[i]});
	}
	for (int i = 0; i < n; i++) {
		auto it = s.begin();
		if (it->si != a[i]) {
			v.pb(it->si);
			cnt2[it->si]--;
			if (!cnt2[it->si])
				s.erase(it);
			continue;
		}
		++it;
		if (it == s.end())
			break;
		v.pb(it->si);
		cnt2[it->si]--;
		if (!cnt2[it->si])
			s.erase(it);
	}
	if (int(v.size()) == n) {
		cout << "Yes\n";
		for (int i = 0; i < n; i++)
			cout << v[i] << ' ';
		cout << endl;
	} else {
		cout << "No\n";
	}
	return 0;
}