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: Accepted)

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define endl "\n"
#define fast ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
ll M=1e9+7;
const ll mxx=1e5+7;


int main()
{

    int n;
    cin>>n;
    int a[n],b[n];
    for(auto &x:a)cin>>x;
    for(auto &x:b)cin>>x;

    reverse(b,b+n);
    
    int l=-1,r=-1;
    for(int i=0;i<n;i++)
    {
       if(a[i]==b[i])
       {
       	if(l==-1)l=i,r=i;
       	else r=i;
       }
    }
    vector<int> v;
    if(l!=-1)
    {
        int num=a[l];
        for(int i=0;i<l;i++)
        {
        	 if(a[i]!=num && b[i]!=num)v.push_back(i);
        }
         for(int i=r+1;i<n;i++)
        {
        	 if(a[i]!=num && b[i]!=num)v.push_back(i);
        }
        if(v.size()<r-l+1){cout<<"No";return 0;}
        int cur=l;
        while(cur<=r)
        {
            swap(b[cur],b[v.back()]);
            v.pop_back();
            cur++;
        }
    }
    cout<<"Yes"<<endl;
    for(auto &x:b)cout<<x<<" ";
}