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: Runtime Error)

#include <bits/stdc++.h>
#include <string>
#include <cmath>
#include <sstream>
#define FOR(i, a, b) for (int i = (a); i < (b); i++)
#define FORD(i, a, b) for(int i = (a); i >=(b); i--)
#define pb(a) push_back(a)
#define mp(a,b) make_pair(a,b)
typedef long long int ll;
using namespace std;
ll mod1=1000000007;
ll mod2=67280421310721;
ll mod3=998244353;
ll INF=1e18;

ll binpow1(ll a,ll b){
    ll res=1;
    while (b>0){
        if (b&1)
        res=res*a;
        a=a*a;
        b>>=1;
    }
    return res;
}

ll binpow2(ll a,ll b,ll m){
    a%=m;
    ll res=1;
    while (b>0){
        if (b & 1)
        res=res*a%m;
        a=a*a%m;
        b>>=1;
    }
    return res;
}

ll gcd(ll a,ll b){
    if (b==0)
        return a;
    else
        return gcd(b,a%b);
}

map<ll,ll> fact(ll n) {
    map<ll,ll> factorization;ll c=0;
    while (n % 2 == 0) {
        c++;
        n /= 2;
    }
    if(c>0)
    {
        factorization.insert({2,c});
    }c=0;
    for (ll d = 3; d * d <= n; d += 2) {
        while (n % d == 0) {
            c++;
            n /= d;
        }
        if(c>0)
    {
        factorization.insert({d,c});
    }c=0;
    }
    if (n > 1)
       {
           factorization.insert({n,1});
       }
    return factorization;
}

ll func1(ll n, ll r, ll p)
{
    // Optimization for the cases when r is large
    if (r > n - r)
        {r = n - r;}

    // The array C is going to store last row of
    // pascal triangle at the end. And last entry
    // of last row is nCr
    ll C[r + 1];
    memset(C, 0, sizeof(C));

    C[0] = 1; // Top row of Pascal Triangle

    // One by constructs remaining rows of Pascal
    // Triangle from top to bottom
    for (ll i = 1; i <= n; i++) {

        // Fill entries of current row using previous
        // row values
        for (ll j = min(i, r); j > 0; j--)

            // nCj = (n-1)Cj + (n-1)C(j-1);
           {
               C[j] = (C[j] + C[j - 1]) % p;
           }
    }
    return C[r];
}


int main()
{ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	cout.tie(NULL);
    ll n;
    cin>>n;
    ll a[n];ll b[n];ll f=1;
    FOR(i,0,n)
    {
        cin>>a[i];
    }
    FOR(i,0,n)
    {
        cin>>b[i];
    }
    ll c[10]={0};
    ll d[10]={0};
    FOR(i,0,n)
    {
        c[a[i]]++;
        d[b[i]]++;
    }
    FOR(i,0,10)
    {
        if(d[i]>(n-c[i]))
        {
            f=0;break;
        }
    }
    if(f==1)
    {
        cout<<"Yes"<<'\n';
        FOR(i,0,n)
        {
            cout<<b[i]<<" ";
        }
        cout<<'\n';
    }
    else
    {
        cout<<"No"<<'\n';
    }



	return 0;
}