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>
#define IOS {ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);}
#define endl "\n"
#define float long double
#define int long long int
#define f(i, a, b) for(int i=a; i<b; i++)
#define fb(i, a, b) for(int i=a; i>=b; i--)
#define all(v) (v).begin(), (v).end()
#define pb push_back
#define fi first
#define se second
#define vi vector <int>
#define pi pair <int, int>
#define LB lower_bound
#define UB upper_bound
#define g(t, n) get<n>(t)
#define fp(x, y) fixed<<setprecision(y)<<x
#define PI 3.141592653589793238462643383279502884
using namespace std;

const int M = (int)2e5+5;
const int mod = (int)1e9+7;

void solve()
{
    int n;
    cin>>n;
    int a[n], b[n], cnt[n+1] = {};
    f(i, 0, n) cin>>a[i], cnt[a[i]]++;
    f(i, 0, n) cin>>b[i], cnt[b[i]]++;
    f(i, 0, n+1)
    {
    	if(cnt[i] > n)
    	{
    		cout<<"No\n";
    		return;
    	}
    }
    cout<<"Yes\n";
    reverse(b, b+n);
    int s = 0, e = n-1;
    f(i, 0, n)
    {
    	if(b[i] == a[i])
    	{
    		if(b[e] != b[i]) swap(b[i], b[e--]);
    		else swap(b[i], b[s++]);
    	}
    }
    f(i, 0, n) cout<<b[i]<<' ';
    cout<<endl;
}

int32_t main()
{
    IOS
    int t = 1;
    // cin>>t;
    f(tno, 1, t+1)
    {
        // cout<<"Case #"<<tno<<": ";
        solve();
    }
    return 0;
}