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;
const int maxn=2e5+10;
int n;
int a[maxn],b[maxn];
int main()
{
	ios::sync_with_stdio(false);cin.tie(0);
	cin>>n;
	for (int i=1;i<=n;i++) cin>>a[i];
	for (int i=1;i<=n;i++) cin>>b[i];
	reverse(b+1,b+n+1);
	int l=0,r=0;
	for (int i=1;i<=n;i++)
	{
		if (a[i]!=b[i]) continue;
		l=i;
		while (a[i]==b[i] && i<=n) i++;
		r=i-1;
	}
	if (l==0)
	{
		cout<<"Yes\n";
		for (int i=1;i<=n;i++) cout<<b[i]<<' ';cout<<'\n';
		return 0;
	}
//	cout<<l<<' '<<r<<endl;
	int num=l;
	for (int i=1;i<l && num<=r;i++)
	{
		swap(b[i],b[num]);
		num++;
	}
	for (int i=n;i>r && num<=r;i--)
	{
		swap(b[i],b[num]);
		num++;
	}
	bool bo=1;
	for (int i=1;i<=n;i++)
		if (a[i]==b[i]) {bo=0;break;}
	if (!bo) {cout<<"No\n";return 0;}
	cout<<"Yes\n";
	for (int i=1;i<=n;i++) cout<<b[i]<<' ';cout<<'\n';
	return 0;
}