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: Time Limit Exceeded)

#include <map>
#include <queue>
#include <cmath>
#include <ctime>
#include <vector>
#include <cstdio>
#include <complex>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <tr1/unordered_map>
#define mo 1000000007
#define num(x) (x>='0' && x<='9')
typedef unsigned long long ull;
typedef long long ll;
using namespace std::tr1;
using namespace std;
int read(){
    int p=0, q=1;
    char ch=getchar();
    while (!num(ch))
        (ch=='-'?q=-1:0),
        ch=getchar();
    while (num(ch)) p=p*10+ch-'0', ch=getchar();
    return p*q;
}
int power(ll a, int b){
	ll c=1;
	while (b){
		if (b&1) c=c*a%mo;
		a=a*a%mo;
		b>>=1;
	}
	return c;
}
#define N 200005
int n, a[N], b[N], f[N], g[N], ans[N], flag[N];
int main(){
	n=read();
	for (int i=1;i<=n;i++) a[i]=read(), f[a[i]]++;
	for (int i=1;i<=n;i++) b[i]=read(), g[b[i]]++;
	for (int i=1;i<=n;i++)
		if (f[i]+g[i]>n){
			puts("No");
			return 0;
		}
	puts("Yes");
	for (int i=1, j=1, k=1;i<=n;i++){
		if (!f[b[i]]) continue;
		while (j<=n && a[j]<=b[i]) j++;
		if (j<=n) ans[j]=b[i], flag[j++]=1;
		else{
			while (a[k]==b[i] || flag[k])
				if (++k>n) k=1;
			ans[k]=b[i], flag[k++]=1;
			if (k>n) k=1;
		}
	}
	for (int i=1, j=1;i<=n;i++)
		if (!ans[i]){
			while (f[b[j]] || !g[b[j]]) j++;
			g[b[j]]--;
			ans[i]=b[j];
		}
	for (int i=1;i<=n;i++) printf("%d%s", ans[i], i==n?"\n":" ");	
    return 0;
}
/*
5
1 1 2 2 3
2 2 3 3 3
*/