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>
int main(){
  int n,a[200010]={},b[200010]={},c[200010],d[200020],e=0,i,g[200020],f=0,h[200010],r,y,yy,z;
  scanf("%d",&n);
  for(i=0;i<n;i++){
    scanf("%d",&c[i]);
    a[c[i]]++;
  }
  for(i=0;i<n;i++){
    scanf("%d",&d[i]);
    b[d[i]]++;
  }
  for(i=0;i<n;i++){
    if(a[i]+b[i]>n){
      printf("No\n");
      return 0;
    }
  }
  printf("Yes\n");
  f=-1;
  for(i=0;i<n;i++)g[i]=d[n-i-1];
  //for(i=0;i<n;i++)printf("%d ",g[i]);printf("\n");
  for(i=0;i<n;i++){
    if(c[i]==g[i]&&f==-1)f=i;
  }
  //printf("%d %d\n",f,g[f]);
  for(i=0;i<n;i++)h[i]=g[f];
  
  if(f==-1){
    for(i=0;i<n;i++)h[i]=g[i];
  }
  else{
    y=b[g[f]];
    if(y>f)y=f;
    yy=b[g[f]]-y;
    z=n-f-b[g[f]];
    //printf("%d %d\n",y,yy);
    for(i=0;i<f;i++)h[i+y]=g[i];
    for(i=f+y+yy;i<n;i++)h[i-yy]=g[i];
  }
  for(i=0;i<n;i++){
    if(c[i]==h[i]){
      printf("No\n");
      return 0;
    }
  }
  for(i=0;i<n;i++)printf("%d%c",h[i],i+1==n?'\n':' ');
  return 0;
}