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;
typedef long long int ll;
#define INF 1e9+7
#define rep(i,n) for(ll i=0;i<n;i++)
#define pii pair<int,int>
#define pll pair<ll,ll>
const int maxn = 200000;
ll mod = INF;

int n;
int a[maxn],b[maxn];
vector<int> ans;

int main(){

    cin >> n;
    rep(i,n){
        cin >> a[i];
    }

    rep(i,n){
        cin >> b[i];
    }
    int idx=n-1;
    rep(i,n){
        if(a[i]!=b[n-1-i]){
            ans.push_back(b[n-1-i]);
        }else{
            break;
        }
        idx--;
    }
    int i=0;
    while(ans.size()!=n){
        ans.push_back(b[i]);
        i++;
    }

    rep(i,n){
        cout << ans[i]<<endl;
    }
    
    rep(i,n){
        if(a[i]==ans[i]){
            cout << "No"<<endl;
            return 0;
        }
    }

    cout << "Yes"<<endl;
    
    return 0;
}