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 rep(i,n) for (int i = 0; i < (n); ++i)
#define repit(it, li) for(auto it=li.begin(); it!=li.end(); it++)
using namespace std;
using ll = long long;
using P = pair<int,int>;

int main(){
	int n;
	cin>>n;
	vector<int> a(n), b(n);
	rep(i, n) cin>>a[i];
	rep(i, n) cin>>b[i];

	map<int, int> ma;
	rep(i, n) ma[a[i]]++;
	rep(i, n) ma[b[i]]++;
	bool ok=true;
	for(auto p : ma) if(p.second>n) ok=false;
	if(!ok){
		cout<<"No"<<endl;
		return 0;
	}

	cout<<"Yes"<<endl;
	vector<bool> ac(n, true);
	int l=0, r=0, c=0, d=0;
	for(int i=1; i<n; i++){
		l=r;
		for(; a[r]==i; r++);
		c=d;
		for(; b[d]==i; d++);
		if(l<r && c<d){
			for(int i=(l-(d-1)+n)%n; (c+i)%n!=r%n; i=(i+1)%n) ac[i]=false;
		}
	}

	int e;
	rep(i, n) if(ac[i]) e=i;

	rep(i, n) cout<<b[(-e+i+n)%n]<<" ";
	cout<<endl;
	return 0;
}