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 (Clang 10.0.0) (judge verdict: Runtime Error)

#include<stdio.h>

int main(){
  int n,amax=0,bmax=0;
  scanf("%d",&n);
  int a[n+1],b[n+1],sum[n+1],pointa[n+1],pointb[n+1];
  for(int i=1;i<=n;i++){
    sum[i]=0;
    pointa[i]=0;
    pointb[i]=0;
  }
  for(int i=1;i<=n;i++){
    scanf("%d",&a[i]);
    sum[a[i]]+=1;
    if(a[i]!=a[i-1]){
      pointa[a[i-1]]=i-1;
    }
  }
  for(int i=1;i<=n;i++){
    scanf("%d",&b[i]);
    sum[b[i]]+=1;
    if(b[i]!=b[i-1]){
      pointb[b[i-1]]=i-1;
    }
  }
  amax=a[n];
  bmax=b[n];

  for(int i=1;i<=n;i++){
    if(sum[i]>n){
      printf("No");
      return 0;
    }
  }
  printf("Yes\n");
  
  for(int i=1;i<=n;i++){
    if(a[i]==b[i]){
      if(a[i]!=bmax){
        int s=b[i];
        b[i]=b[pointb[a[i]]+1];
        b[pointb[a[i]]+1]=s;
        pointb[a[i]]+=1;
      }else{
        for(int j=1;j<i;j++){
          if(b[j]!=b[i] && a[j]!=b[i]){
            int t=b[i];
            b[i]=b[j];
            b[j]=t;
            break;
          }
        }
      }
    }
  }
  
  for(int i=1;i<=n;i++){
    printf("%d ",b[i]);
  }
  return 0;
}