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 pi acos(-1.0)
const int MOD=1e9+7;
#define pb push_back
#define cy cout<<"Yes\n"
#define cn cout<<"No\n"
#define ci cout<<"impossible\n"
typedef unsigned long long int ull;
typedef long long int ll;
const int MAX=500005;
ll ar[MAX],br[MAX];
const int M=1e4;
ll cnt[M];
int main()
{
    int n;
    cin>>n;
    for(int i=0;i<n;i++)cin>>ar[i];
    for(int i=0;i<n;i++)cin>>br[i];
    reverse(br,br+n);
    int l=0,r=n-1;
    for(int i=0;i<n;i++)
    {
        if(ar[i]==br[i])
        {
            if(br[l]!=br[i] and ar[l]!=br[i])
            {
                swap(br[l],br[i]);
                l++;
            }
            else if(br[r]!=br[i] and ar[r]!=br[i])
            {
                swap(br[r],br[i]);r--;
            }
            else
            {
                cn;
                return 0;
            }
        }
    }
    bool ok=1;
    for(int i=0;i<n;i++)
    {
        if(ar[i]==br[i])
        {
            cn;
            return 0;
        }
    }
    cy;
    for(int i=0;i<n;i++)
        cout<<br[i]<<" ";

}