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>

using namespace std;

const int N = 2e5+5,mod=1e9+7;
int freq1[N],freq2[N],taken[N];
int main() {
    int n;
    cin>>n;
    int arr1[n],arr2[n];
    std::vector<int> v;
    for (int i = 0; i < n; ++i)
    {
   		cin>>arr1[i];
   		freq1[arr1[i]]++;
    }
    for (int i = 0; i < n; ++i)
    {
   		cin>>arr2[i];
   		freq2[arr2[i]]++;
    }
    set<int> ss;
    ss.insert(arr1,arr1+n);
    int l=0,r=n-1;
    for (auto it = ss.begin(); it != ss.end()&&l<r; ++it)
    {
    	while(l<n&&l<=r){
    		if(arr2[l]==*it||!freq1[*it])
    			break;
    		v.push_back(arr2[l]);
    		freq1[*it]--;
    		l++;
    	}
    	while(r>=0&&l<r){
    		if(arr2[r]==*it||!freq1[*it])
    			break;
    		v.push_back(arr2[r]);
    		freq1[*it]--;
    		r--;
    	}
    	if(freq1[*it]){
    		cout<<"No";
    		return 0;
    	}
    }
    if(v.size()!=n){
    	cout<<"No";
    	return 0;
    }
    cout<<"Yes\n";
    for (int i = 0; i < n; ++i)
    {
    	cout<<v[i]<<' ';
    }


}