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 N;
    cin >> N;
    vector<int>A(N),B(N);
    for(int i = 0; i < N; i++) cin >> A[i];
    for(int i = 0; i < N; i++) cin >> B[i];
    reverse(B.begin(),B.end());
    for(int i = 0; i < N; i++) {
        if(A[i] == B[i]) {
            int now = N-1;
            for(int j = i+1; j < N; j++) if(A[i] != B[j]) {now = j;break;}
            if(!(N-now < now-i+1)) {
                for(int j = i; j < now; j++) swap(B[j],B[now+j-i]);
                i = now*2-i-1;
            }
            else {
                int old = now;
                for(int j = i-1; j >= 0; j--) if(A[i] != B[j]) {now = j;break;}
                if(!(now == -1 || now+1 < old-now+1)) {
                    for(int j = i; j > now; j--) swap(B[j],B[now+j-i]);
                    if(old != -1) i = old-1;
                }
                else {
                    cout << "No" << endl;
                    return 0;
                }
            }
        }
    }
    cout << "Yes" << endl;
    for(int i = 0; i < N; i++) {
        cout << B[i] << ((i+1 == N)?"\n":" ");
    }
}