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>
using namespace std;
typedef long long int ll;
typedef pair<ll,ll> P;
typedef vector<ll> VI;
typedef vector<VI> VVI;
const ll MOD = 1000000007;
const ll INF = 1e18;
#define REP(i,n) for(int i=0;i<n;i++)
#define ALL(v) v.begin(),v.end()

struct lr{
	int l;
	int r;
};

int main(){
	int n; cin >> n;
	VI a(n), b(n);
	REP(i,n) cin >> a[i];
	REP(i,n) cin >> b[i];
	vector<lr> ap(n+1,{-1,-1}), bp(n+1,{-1,-1});
	ap[a[0]].l=0; ap[a[n-1]].r=n-1;
	REP(i,n-1){
		if(a[i]!=a[i+1]){
			ap[a[i]].r=i;
			ap[a[i+1]].l=i+1;
		}
	}
	bp[b[0]].l=0; bp[b[n-1]].r=n-1;
	REP(i,n-1){
		if(b[i]!=b[i+1]){
			bp[b[i]].r=i;
			bp[b[i+1]].l=i+1;
		}
	}
	VI s(n,0);
	for(int i=1;i<=n;i++){
		if(ap[i].l!=-1&&bp[i].l!=-1){
			int al=ap[i].r-ap[i].l+1;
			int bl=bp[i].r-bp[i].l+1;
			int sl=ap[i].l-bl+1-bp[i].l;
			int sr=ap[i].r+bl-1-bp[i].r+1;
			if(sl<0){
				s[(sl+n)%n]++;
				s[0]++;
			}
			else{
				s[sl]++;
			}
			if(sr>=n){
				s[0]++;
				s[sr%n]--;
			}
			else{
				s[sr]--;
			}
		}
	}
	int p=0;
	REP(i,n){
		p+=s[i];
		if(p==0){
			cout << "Yes" << endl;
			REP(j,n)
				cout << b[(j+i)%n] << " ";
			cout << endl;
			return 0;
		}
	}
	cout << "No" << endl;
	return 0;
}