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)

#define IO ios::sync_with_stdio(false);cin.tie();cout.tie(0)
#pragma GCC optimize(2)
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long ll;
const int N=200010;
const ll mod=1e9+7;
int a[N],b[N],n;
int cnta[N],cntb[N];
int ans[N];

int main()
{
    IO;
    int T=1;
    //cin>>T;
    while(T--)
    {
        cin>>n;
        for(int i=1;i<=n;i++) 
        {
            cin>>a[i];
            cnta[a[i]]++;
        }
        for(int i=1;i<=n;i++)
        {
            cin>>b[i];
            cntb[b[i]]++;
        }
        for(int i=1;i<=n;i++)
            if(cnta[i]+cntb[i]>n) return cout<<"No\n",0;
        reverse(b+1,b+1+n);
        vector<int> p,q;
        int now=0,cnt=0;
        for(int i=1;i<=n;i++)
        {
            if(a[i]!=b[i]) continue;
            cnt++;
            p.push_back(i);
            now=b[i];
        }
        for(int i=1;i<=n;i++)
        {
            if(!cnt) break;
            if(a[i]==b[i]) continue;
            if(a[i]!=now&&b[i]!=now)
            {
                cnt--;
                q.push_back(i);
            }
        }
        for(int i=0;i<q.size();i++) swap(b[p[i]],b[q[i]]);
        for(int i=1;i<=n;i++) cout<<b[i]<<' ';
        cout<<'\n';
    }
    return 0;
}