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 <stdio.h>
#include <string.h>
#include <stdlib.h>
 
int main() {
    int n;
    scanf("%d", &n);
    int a[n], b[n];
    int i, j;
    int temp = 0;
    int check = 0;
    for(i = 0; i < n; i ++){
    scanf("%d", &a[i]);
    }
    for(i = 0; i < n; i ++){
    scanf("%d", &b[n - i - 1]);
    }
    for (i = 0; i < n; i ++) {
        if (a[i] == b[i]) {
            for (j = 0; j < i; j ++) {
                if(a[j] != b[i] && a[i] != b[j]){
                    temp = b[i];
                    b[i] = b[j];
                    b[j] = temp;
                    break;
                }
                if(j == n -1){
                    check = -1;
                }
            }
        }
    }
    if(check == 0){
        printf("Yes\n");
        for (i = 0; i < n; i ++) {
            printf("%d ", b[i]);
        }
    }else{
        printf("No\n");
    }
    return 0;
}