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 int long long

signed main(){
    // cout << fixed << setprecision(10) << flush;

    int n;
    cin >> n;
    vector<int> a(n), b(n);
    vector<int> numa(n+1, 0), numb(n+1, 0);
    for(int i=0; i<n; i++){
        cin >> a[i];
        numa[a[i]]++;
    }
    for(int i=0; i<n; i++){
        cin >> b[i];
        numb[b[i]]++;
    }
    reverse(b.begin(), b.end());

    for(int i=0; i<n; i++){
        if(a[i] == b[i]){
            int j;
            if(numa[a[i]] + numb[b[i]] > n){
                cout << "No" << endl;
                return 0;
            }
            if(i % 2 == 0){
                for(j=0; j<n; j++){
                    if(i == j) continue;
                    if(a[i] != a[j] && a[i] != b[j]){
                        swap(b[i], b[j]);
                        break;
                    }
                }
            }
            else{
                for(j=n-1; j>=0; j--){
                    if(i == j) continue;
                    if(a[i] != a[j] && a[i] != b[j]){
                        swap(b[i], b[j]);
                        break;
                    }
                }
            }
        }
    }
    cout << "Yes" << endl;
    for(int i=0; i<n-1; i++){
        cout << b[i] << " ";
    }
    cout << b[n-1] << endl;

    return 0;
}