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;
typedef long long ll;
#define FIO ios_base::sync_with_stdio(0);cin.tie(0);

const int N = 2e5+5, M = 1e6+6, OO = 0x3f3f3f3f;
int tc;

int main()
{
    FIO
    int n; cin>>n;
    vector<int> a(n), b(n);
    for(auto &it : a)
        cin>>it;
    for(auto &it :b)
        cin>>it;
    reverse(b.begin(), b.end());
    int lp = -1;
    for (int i = 0; i < n; i++)
    {
        if(a[i] == b[i])
        {
            lp = i;
            break;
        }
    }
    int rp = n;
    for(int i = n-1 ; i != -1; i-- )
    {
        if(a[i] == b[i])
        {
            rp = i;
            break;
        }
    }
    if(lp == -1)
    {
        cout<<"Yes"<<endl;
        for(auto x : b)
            cout<<x<<" ";
        return 0;
    }
    for(int i = 0 ; i< n ; i++)
    {
        if(b[i]!= b[lp] && b[lp] != a[i] && lp <= rp)
        {
            swap(b[i], b[lp]);
            lp++;
        }
    }
    if(lp <= rp)
       return  cout<<"No"<<endl, 0;
    cout<<"Yes"<<endl;
        for(auto x : b)
            cout<<x<<" ";
    
    
    return 0;
}