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)

#pragma GCC optimize ("O3")
#include <iostream>
#include <iomanip>
#include <istream>
#include <ostream>
#include <sstream>
#include <iterator>
#include <vector>
#include <algorithm>
#include <queue>
#include <deque>
#include <list>
#include <stack>
#include <map>
#include <unordered_map>
#include <set>
#include <bitset>
#include <utility>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <string>
#include <ctime>
#include <cctype>
#include <cstdlib>
#include <numeric>
#define IINF 1000000000
#define INF 3223372036854775807
#define MOD 1000000007
#define mod 1000000007
#define INT_MAX_ 2147483647
#define EPS (1e-10)
#define REP(i, a, n) fo-r (ll i = a; i < (ll)(n); i++)
#define REPE(i, a, n) for (ll i = a; i <= (ll)(n); i++)
//#define rep(i,n)for (ll i = 0; i < (ll)(n); i++)
#define rep(i,l,r)for(ll i=(l);i<(r);i++)
#define rep1(i,n) for(int i=1;i<=(int)(n);i++)
#define Endl endl
#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define mt make_tuple
#define eb emplace_back
#define mmax(x,y)(x>y?x:y)
#define mmin(x,y)(x<y?x:y)
#define chmax(x,y) x=mmax(x,y)
#define chmin(x,y) x=mmin(x,y)
#define all(x) (x).begin(),(x).end()
#define siz(x) (ll)(x).size()
#define PI acos(-1.0)
#define me memset
#define bit(n,k) ((n>>k)&1)
#define lg length()
using namespace std;
typedef long long int ll;
typedef unsigned long long int ull;
typedef long double ld;
typedef pair<int,int>Pin;
typedef pair<ll,ll>Pll;
template<class T> using V=vector<T>;
template<typename T> using min_priority_queue = priority_queue<T, vector<T>, greater<T> >;
long long GCD(long long a, long long b) {return b?GCD(b,a%b):a;}
long long LCM(long long a, long long b) {return a/GCD(a,b)*b;}
ll pom(ll a,ll n,int m){ll x=1;for(a%=m;n;n/=2)n&1?x=x*a%m:0,a=a*a%m;return x;}
#define invp(a,p)pom(a,p-2,p)
int dx[4]={-1,0,1,0};
int dy[4]={0,-1,0,1};
int ddx[8]={-1,0,1,0,1,1,-1,-1};
int ddy[8]={0,-1,0,1,1,-1,1,-1};
ll cmp1(pair<Pll,ll> a,pair<Pll,ll> b){
        return a.fi.se>b.fi.se;
}
ll cmp2(pair<ll,min_priority_queue<ll>> a,pair<ll,min_priority_queue<ll>> b){
    return a.fi > b.fi;
}
//----------------------------------------------------------------------

//----------------------------------------------------------------------
int main(int argc, char * argv[]){
    cin.tie(0);
    ios::sync_with_stdio(false);
    //-------------------------------  
    //ll begin_t=clock();
    //freopen("big.txt", "r", stdin);
    //freopen("out3.txt", "w", stdout);
    //------------------------------
    ll n;cin>>n;
    V<ll>a(n),b(n);
    for(ll i=0;i<n;i++)cin>>a[i];
    for(ll i=0;i<n;i++)cin>>b[i];
    deque<ll>Q;
    reverse(all(a));
    reverse(all(b));
    for(ll i=0;i<n;i++)Q.push_back(b[i]);
    V<ll>ans;
    for(ll i=0;i<n;i++){
        ll q1 = Q.front();Q.pop_front();
        ll q2=a[i];
        bool abc=0;
        if(Q.size()>=1){
            q2=Q.back();
            Q.pop_back();
            abc=1;
        }
        bool kkt=0;
        if(q1!=a[i]){
            ans.pb(q1);
            if(abc==1){
                Q.push_back(q2);
            }
            kkt=1;
        }
        else{
            if(q2!=a[i]&&abc==1){
                kkt=1;
                Q.push_front(q1);
                ans.pb(q2);
            }
        }
        if(kkt==0){
            cout<<"No"<<endl;
            return 0;
        }
    }
    for(ll i=0;i<n;i++){
        if(a[i]==ans[i]){
            cout<<"No"<<endl;
            return 0;
        }
    }
    if(ans.size()!=n){
        cout<<"No"<<endl;
        return 0;
    }
    cout<<"Yes"<<endl;
    for(ll i=0;i<n;i++){
        cout<<ans[i]<<" ";
    }
    cout<<endl;
    //------------------------------
    //fclose(stdin);
    //fclose(stdout);
    //ll end_t=clock();cout<<"time="<<end_t-begin_t<<"ms"<<endl;
    //------------------------------- 
    return 0;
}
//----------------------------------------------------------------------