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)

/*
   Username : moondancer
   Name : Moon Dancer
*/


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

#define int long long

signed main(){
    ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
    int n;
    cin>>n;
    int a[n];
    for(int i=0; i<n; ++i)
        cin>>a[i];
    int b[5*n];
    for(int i=0; i<n; ++i){
        cin>>b[i];
        b[i+n]=b[i];
        b[i+2*n]=b[i];
        b[i+3*n]=b[i];
        b[i+4*n]=b[i];
    }
    int ans[n];
    memset(ans,-1,sizeof(ans));
    int l=0;
    for(int i=0; i<n; ++i){
        while(l<5*n && (b[l] == a[i] || b[l] == -1))
            l++;
        if(l==5*n) break;
        if(b[l] != a[i] && b[l] != -1){
            ans[i]=b[l];
            int x=l%n;
            b[x]=b[x+n]=b[x+2*n]=b[x+3*n]=b[x+4*n]=-1;
            l++;
        }
        else
            break;
    }
    for(int i=0; i<n; ++i){
        if(ans[i] == -1){
            cout<<"No";
            return 0;
        }
    }
    cout<<"Yes\n";
    for(int i=0; i<n; ++i){
        cout<<ans[i]<<" ";
    }
    return 0;
}