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)

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

int main() {
    int tc = 1;
    while(tc--) {
        int n;
        cin>>n;
        vector<int> a(n), b(n);
        vector<int> c1(n+1, 0);
        int mx = 0;
        for(int i =0; i<n; i++) {
            cin>>a[i];
            c1[a[i]]++; 
        }
        set<int> unused;
        for(int i =0; i<n; i++) {
            cin>>b[i];
            c1[b[i]]++;
            unused.insert(i);
        }
        for(int i = 1; i<=n ; i++) {
            if(c1[i] > n) {
                cout<<"No"<<endl;
                return 0;
            }
        }
        cout<<"Yes"<<endl;
        vector<int> res(n, 0);
        
        auto itr = unused.begin();
        for(int i = 0; i<n; i++) {
            itr = prev(unused.end());
            if(a[*itr] == b[i])
                itr = unused.begin();
            while(itr != unused.end() && a[*itr] == b[i])
                itr++;
            res[*itr] = b[i];
            unused.erase(*itr);
        }
        for(int i=0; i<n; i++)
            cout<<res[i]<<" ";
    }
    return 0;
}