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<bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp> 
#include <ext/pb_ds/tree_policy.hpp>
#define ll long long int
#define ld long double
#define pb push_back
#define fi first
#define se second
#define all(x) x.begin(),x.end()
#define mem(x,y) memset(x,y,sizeof(x))
#define pii pair<int,int>
#define pll pair<ll,ll>
#define INF 1e9
#define INFL 1e18
#define mod 1000000007
//#define mod 998244353
#define fast ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
using namespace std;
using namespace __gnu_pbds; 
typedef tree<int,null_type,less<int>,rb_tree_tag,tree_order_statistics_node_update> os;
typedef tree<pii,null_type,less<pii>,rb_tree_tag,tree_order_statistics_node_update> os_pair;

ll power(ll x,ll n){ll res =1;while(n>0){if(n%2==1){res=res*x;}x=x*x;n=n/2;}return res;}
ll powm(ll a,ll b) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}

//cout<< fixed << setprecision(10)
//__builtin_popcountll()



int main(){

    fast;
    int n,i,j;
    cin>>n;
    vector<int>a(n),b(n);
    for(i=0;i<n;i++)
    cin>>a[i];
    for(i=0;i<n;i++)
    cin>>b[i];
    reverse(all(b));
    vector<int>v;
    for(i=0;i<n;i++)
    {
    	if(a[i]==b[i])
    	v.pb(i);
	}
	if(v.size()!=0)
	{
		int l=v[0];
		int r=v[(int)v.size()-1];
		int x=a[l];
		for(i=0;i<n;i++)
		{
			if(a[i]!=x&&b[i]!=x&&l<=r)
			{
				swap(b[i],a[i]);
				l++;
			}
		}
		if(l<=r)
		{
			cout<<"No";
			return 0;
		}
	}
	cout<<"Yes\n";
	for(i=0;i<n;i++)
	{
		cout<<b[i]<<" ";
	}
    
    
    

    

}