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: Accepted)

#include<bits/stdc++.h>
using namespace std;

#define rep(i,n) for(int i=0;i<n;i++)
 
int main()
{

	int n,flag=0;
    cin >> n;
    int a[n],b[n],ans[n];
    
    rep(i,n) cin >> a[i];
    rep(i,n) cin >> b[i];
    rep(i,n){
        ans[i]=b[n-1-i];  
    }
    int l=0,r=n-1;
    rep(i,n){
        if(a[i]==ans[i]){
            if(a[i]!=ans[l] && a[l]!=ans[i]){
                swap(ans[i],ans[l]);
                l++;
            }
            else if(a[i]!=ans[r] && a[r]!=ans[i]){
                swap(ans[i],ans[r]);
                r--;
            }
            else
                flag=1;
               
            
        }
    }
    if(flag==1) cout << "No" << endl;
    else
    {
    cout << "Yes" << endl;
    rep(i,n) cout << ans[i] << " ";
    cout << "" << endl;
	}
}