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("Ofast") 
#include<bits/stdc++.h>
using namespace std;
#define pb push_back
#define fi first
#define se second
#define sz(a) a.size()
#define all(a) a.begin(),a.end()
#define lb lower_bound
#define ub upper_bound
#define owo ios_base::sync_with_stdio(0);cin.tie(0);
#define MOD (ll)(1e9+7)
#define INF (ll)(1e18)
#define debug(...) fprintf(stderr, __VA_ARGS__),fflush(stderr)
#define time__(d) for(long blockTime = 0; (blockTime == 0 ? (blockTime=clock()) != 0 : false);\
debug("%s time : %.4fs\n", d, (double)(clock() - blockTime) / CLOCKS_PER_SEC))
typedef long long int ll;
typedef long double ld;
typedef pair<ll,ll> PII;
typedef pair<int,int> pii;
typedef vector<vector<int>> vii;
typedef vector<vector<ll>> VII;
ll gcd(ll a,ll b){if(!b)return a;else return gcd(b,a%b);}
int main()
{
	int n;
	cin>>n;
	deque<ll>a(n),b(n);
	vector<ll>cnt(n+1);
	bool ok = 1;
	for(int i=0;i<n;i++){
		cin>>a[i];
		cnt[a[i]]++;
		if(cnt[a[i]] > n/2+1)ok = 0;
	}
	for(int i=1;i<=n;i++)cnt[i] = 0;
	for(int i=0;i<n;i++){
		cin>>b[i];
		cnt[b[i]]++;
		if(cnt[b[i]] > n/2+1)ok = 0;		
	}
	if(!ok)cout<<"No";
	else{
		cout<<"Yes"<<'\n';
		vector<ll>ansa(n),ansb(n);
		int cnt = 0;
		while(!a.empty() && !b.empty()){
			if(a.front()!=b.front()){
				ansa[cnt] = a.front();
				ansb[cnt] = b.front();
				cnt++;
				a.pop_front();
				b.pop_front();
			}else{
				b.pb(b.front());
				b.pop_front();
			}
		}
		for(int i=0;i<n;i++)cout<<ansa[i]<<" ";
		cout<<'\n';
		for(int i=0;i<n;i++)cout<<ansb[i]<<" ";
	}
}