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)

/// Bismillahir Rahmanir Rahim
//Author: Tanvir Hussain
//ICE,NSTU
#include<bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
using namespace std;
const long long MOD = 1000000007;
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
#define SET(x) memset(x, 0, sizeof(x))
#define SET2d(x,m,n) memset(x, 0, sizeof(x[0][0]) * m * n)
#define SETBOOL(x) memset(x,false,sizeof(x))
#define CLR(x) memset(x, -1, sizeof(x))
#define CLR2d(x,m,n) memset(x, -1, sizeof(x[0][0]) * m * n)
#define mp make_pair
#define PII pair<int, int>
#define pf printf
#define sf scanf
#define ALL(x) x.begin(),x.end()
#define pb push_back
#define IOS ios::sync_with_stdio(false); cin.tie(0);
#define np std::string::npos
#define highest(x) numeric_limits<x>::max()
#define lowest(x) numeric_limits<x>::min()
#define Inf INFINITY
#define minv(v) *min_element(v.begin(),v.end())
#define maxv(v) *max_element(v.begin(),v.end())
#define cases(cs,t) for(int cs=1;cs<=t;cs++)
#define PI acos(-1)
#define no1 __builtin_popcount
#define BOUNDARY(i, j) ((i >= 0 && i < row) && (j >= 0 && j < column))
#define uniq(vec) vec.resize(distance(vec.begin(),unique(vec.begin(),vec.end())))
#define ordered_set tree<int, null_type,less<int>, rb_tree_tag,tree_order_statistics_node_update>
#define sz(a) int(a.size())
#define ff first
#define ss second
//#define endl "\n"
#define forch(it,s) for(auto it:s)
#define each(it,s) for(auto it = s.begin(); it != s.end(); ++it)
#define rep(i,a) for(int i=0; i<a;i++)
#define rep1(i,a,b) for(int i=(a);i<=(b);++i)
#define irep(i,b,a) for(int i=(b);i>=(a);--i)
#define bits(n) __builtin_popcount(n)
#define maxpq priority_queue<int>
#define minpq priority_queue<int, vector<int>, greater<int> >



int gcd(int a, int b) { if (a == 0) return b; return gcd(b % a, a);}
typedef long long ll;
typedef vector<int> vi;
typedef vector<ll> vll;
typedef vector<pair<int,int>> vpii;
typedef tree<pair<int, int>,null_type,less<pair<int, int>>,rb_tree_tag,tree_order_statistics_node_update> ordered_multiset;
int dx8[] = {0, 0, 1, 1, 1, -1, -1, -1};
int dy8[] = {1,-1, 1, -1, 0, 0, -1, 1};
int dx4[] = {0, 0, 1, -1};
int dy4[] = {1, -1, 0, 0};
const int maxx=100005;

inline void normal(ll &a) { a %= MOD; (a < 0) && (a += MOD); }
inline ll modMul(ll a, ll b) { a %= MOD, b %= MOD; normal(a), normal(b); return (a*b)%MOD; }
inline ll modAdd(ll a, ll b) { a %= MOD, b %= MOD; normal(a), normal(b); return (a+b)%MOD; }
inline ll modSub(ll a, ll b) { a %= MOD, b %= MOD; normal(a), normal(b); a -= b; normal(a); return a; }
inline ll modPow(ll b, ll p) { ll r = 1; while(p) { if(p&1) r = modMul(r, b); b = modMul(b, b); p >>= 1; } return r; }
inline ll modInverse(ll a) { return modPow(a, MOD-2); }  /// When MOD is prime.
inline ll modDiv(ll a, ll b) { return modMul(a, modInverse(b)); }




//this fuction sorts vector pair according to first element in descending order.
bool sortinrev(const pair<int,int> &a,const pair<int,int> &b)
{
    return a.first>b.first;
}

template<typename T>inline T Bigmod(T base, T power, T MOD){
    T ret=1;
    while(power)
    {
        if(power & 1)ret=(ret*base)%MOD;
        base=(base*base)%MOD;
        power>>=1;
    }
    return ret;
}
double sq(double x) {return x*x;}
ll po(ll b,ll p){
    ll res=1;
    while(p){
        res*=b;
        p--;
    }
    return res;
}
ll lg2(ll x){
    ll res=0;
    while(x>1){
        res++;
        x/=2ll;
    }
    return res;
}
void solve(){
    int n;
    cin>>n;
    vi a(n),b(n);
    rep(i,n){
        cin>>a[i];
    }
    rep(i,n){
        cin>>b[i];
    }
    reverse(ALL(b));
    deque<int> ind;
    int val=-1;
    rep(i,n){
        if(a[i]==b[i]){
            ind.pb(i);
            val=a[i];
        }
    }
    if(val==-1){
        cout<<"Yes"<<endl;
        forch(it,b) cout<<it<<" ";
        cout<<endl;
        return;
    }
    int lb=-1,rb=-1;
    //forch(it,b) cout<<it<<" ";
    //cout<<endl;

    rep(i,n){
        if(a[i]==val){
                if(lb==-1) lb=i;
        rb=i;
        }
    }
    rep(i,n){
        if(b[i]==val){
            lb=min(lb,i);
            rb=max(rb,i);
        }
    }
    lb--;
    rb++;
    //cout<<lb<< " "<<rb<<endl;
    while(ind.size()){
        //cout<<lb<< " "<<rb<<endl;
        if(lb>=0){
            swap(b[ind.front()],b[lb]);
            lb--;
            ind.pop_front();
        }
        else if(rb<n){
            swap(b[ind.front()],b[rb]);
            rb++;
            ind.pop_front();
        }
        else {
            cout<<"No"<<endl;
            return;
        }
    }
    cout<<"Yes"<<endl;
    forch(it,b) cout<<it<<" ";
    cout<<endl;

}

signed main()

{
    IOS;
    /*#ifndef ONLINE_JUDGE
        freopen ("input.txt","r",stdin);
        freopen ("output.txt","w",stdout);
    #endif*/
    int t;
    t=1;
    while(t--){
        solve();
    }

     return 0;

}
///Alhamdulillah