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;
#define rep(i,n) for(int i = 0; i < (int)(n); i++)
#define repr(i,n) for(int i = (int)(n); i >= 0; i--)
#define all(v) v.begin(),v.end()
typedef long long ll;

int main(){
    int N;
    cin >> N;
    vector<int> A(N);
    vector<int> B(N);
    rep(i,N){
        cin >> A[i];
    }
    rep(i,N){
        cin >> B[i];
    }
    map<int,int> mpa;
    map<int,int> mpb;
    rep(i,N){
        mpa[A[i]]++;
        mpb[B[i]]++;
    }
    vector<int> a(N + 1, 0);
    vector<int> b(N + 1, 0);
    rep(i,N){
        a[A[i]] = i + 1;
    }
    rep(i,N){
        b[B[i]] = i + 1;
    }
    int max_sum = 0;
    int max_pointer = 1;
    for (int i = 1; i <= N; i++){
        if (max_sum < mpa[i] + mpb[i]){
            max_sum = mpa[i] + mpb[i];
            max_pointer = i;
        }
    }
    bool possible = true;
    vector<int> C(N);
    if (max_sum > N){
        possible = false;
    }
    else{
        int zurashi = 0;
        rep(i,N){
            zurashi = max(zurashi, a[i + 1] - b[i]);
        }
        rep(i,N){
            C[(i + zurashi) % N] = B[i];
        }
    }
    if (!possible) cout << "No" << endl;
    else{
        cout << "Yes" << endl;
        rep(i,N){
            cout << C[i] << " ";
        }
        cout << endl;
    }
}