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 warning(disable:4996)
#pragma comment(linker, "/STACK:336777216")
#pragma GCC optimize("Ofast")

#include<bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <functional>

using namespace std;
using namespace __gnu_pbds;

typedef long long ll;
#define int long long int
typedef complex<long double> cd;
const long double pi=acos(-1);
typedef double db;
typedef long double ldb;
typedef pair <int, int> pii;
typedef pair <db, db> pdd;
typedef vector<int> vi;
typedef vector<vector<int> > matrix;

#define m1 make_pair
#define pb push_back
#define flush fflush(stdout)
#define IOS std::ios::sync_with_stdio(false); cin.tie(NULL);cout.tie(NULL);
#define reset(x,v) memset(x,v,sizeof(x))

#define fi first
#define se second
#define endl "\n"
#define debug(x) (cerr << #x << ": " << x << "\n")
#define setbits(x) __builtin_popcount(x)
#define setbitsll(x) __builtin_popcountll(x)
#define all(x) x.begin(),x.end()
#define pii pair<int,int>

inline ll gcd(ll a,ll b){if(a==0) return b;return gcd(b%a,a);}
inline ll power(ll a,ll n,ll m){if(n==0)return 1;ll p=power(a,n/2,m);p=(p%m*p%m)%m;if(n%2) return (p%m*a%m)%m;else return p;}
const double EPS = 1e-9;
const ll MOD =998244353;
const ll hell=1000000007;
const int INF = 1e18;
const ll LL_INF = 0x3f3f3f3f3f3f3f3f;
template<typename T>
using ordered_set=tree<T, null_type, less<T>, rb_tree_tag,tree_order_statistics_node_update>;

auto seed = chrono::high_resolution_clock::now().time_since_epoch().count();
std::mt19937 rng(seed);

double startTime;
double getCurrentTime() {
    return ((double)clock() - startTime) / CLOCKS_PER_SEC;
}
const int dx[] = { -1, 0, 1, 0 };
const int dy[] = { 0, 1, 0, -1 };

const int N=200005;

int cnt1[N+1],cnt2[N+1];

void solve()
{
    int n;
    cin>>n;
    int a[n],b[n];
    for(int i=0;i<n;i++)
    {
        cin>>a[i];
        cnt1[a[i]]++;
    }
    for(int i=0;i<n;i++)
    {
        cin>>b[i];
        cnt2[b[i]]++;
    }
    int th=n/2+1;
    set<pair<int,int>> s;
    for(int i=1;i<=N;i++)
    {
        if(cnt1[i]>=th&&cnt2[i]>=th)
        {
            cout<<"No"<<endl;
            return;
        }
        if(cnt2[i])
        {
            s.insert(m1(cnt2[i],i));
        }
    }
    cout<<"Yes"<<endl;
    int ans[n];
    for(int i=0;i<n;i++)
    {
        auto it=(--s.end());
        pair<int,int> p=(*it);
        if(p.se!=a[i])
        {
            cnt2[p.se]--;
            s.erase(it);
            if(cnt2[p.se])
                s.insert(m1(cnt2[p.se],p.se));
            ans[i]=p.se;
        }
        else
        {
            it--;
            p=(*it);
            cnt2[p.se]--;
            s.erase(it);
            if(cnt2[p.se])
                s.insert(m1(cnt2[p.se],p.se));
            ans[i]=p.se;
            
        }
    }
    for(int i=0;i<n;i++)
        cout<<ans[i]<<" ";
    cout<<endl;
    
    
    
}

int32_t main()
{
    IOS;
    
    int t;
    t=1;
    //cin>>t;
    while(t--)
    {
        solve();
    }
    return 0;
}