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 optimization_level 3
#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math,O3")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx")
#include<bits/stdc++.h>

using namespace std;

#define  GODSPEED       ios:: sync_with_stdio(0);cin.tie(0);cout.tie(0);cout<<fixed;cout<<setprecision(15);
#define  f              first                 
#define  s              second
#define  newl           cout<<"\n";
#define  pb             push_back  
#define  mset(a,x)      memset(a,x,sizeof(a))
#define  debv(a)        for(auto it: a)cout<<it<<" ";newl;
#define  deb1(a)        cout<<a<<"\n";
#define  deb2(a,b)      cout<<a<<" "<<b<<"\n";
#define  deb3(a,b,c)    cout<<a<<" "<<b<<" "<<c<<"\n";
#define  deb4(a,b,c,d)  cout<<a<<" "<<b<<" "<<c<<" "<<d<<"\n";
#define  uniq(a)        a.resize(unique(a.begin(), a.end()) - a.begin());
#define  all(a)         a.begin(),a.end()           
 
typedef  long long             ll;
typedef  long double           ld;
typedef  pair<ll,ll>           pll;
typedef  vector<ll>            vll;
typedef  vector<pll>           vpll;

const int  N     =  4e5+5;
const int  mod   =  1e9+7;
const ll   INF   =  0x7f7f7f7f7f7f7f7f;
const int  INFi  =  0x7f7f7f7f;
const int  LEVEL =  log2(N)+1;

ll n, a[N], b[N], cta[N], ctb[N], mx = 0, c[N];

void solve(){
	mset(ctb, -1);
	cin >> n;
	for(int i = 0; i < n; i++) cin >> a[i], cta[a[i]] = i;
	for(int i = 0; i < n; i++){
		cin >> b[i];
		if(ctb[b[i]] == -1) ctb[b[i]] = i;
	}
	for(int i = 1; i <= n; i++){
		mx = max(mx, (n + cta[i] - ctb[i])%n);
	}
	for(int i = 0; i < n; i++){
		c[(i+mx+1)%n] = b[i];
	}
	for(int i = 0; i < n; i++){
		if(c[i] == a[i]){
			deb1("No")
			return;
		}
	}
	deb1("Yes")
	for(int i = 0; i < n; i++) cout << c[i] << " ";
}

int main(){
	GODSPEED;
	int test = 1;
	//cin >> test;
	for(int i = 1; i <= test; i++){
		solve();
	}
	#ifndef ONLINE_JUDGE
	cout<<"\nTime Elapsed: " << 1.0*clock() / CLOCKS_PER_SEC << " sec\n";
	#endif
}