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;

typedef long long ll;
typedef pair <int , int> pii;
typedef long double ld;

const int N = 2e5+10 , mod = 1e9+7;

int cnt[N];
vector <int> pos[N] , pos2[N];

int32_t main(){
   ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);
   int n;
   cin >> n;
   int a[n] , b[n] , ans[n];
   for(int i = 0 ; i < n ; i++)
      cin >> a[i];
   for(int i = 0 ; i < n ; i++){
      cin >> b[i];
      cnt[b[i]]++;
      pos[b[i]].push_back(i);
      pos2[b[i]].push_back(i);
   }
   set <pii> have , rem;
   for(int i = 1 ; i <= n ; i++){
      if(cnt[i]){
         have.insert({cnt[i] , i});
         rem.insert({cnt[i] , i});
      }
   }
   while(have.size()){
      pii x = *have.rbegin();
      auto it = have.end();
      it--;
      have.erase(it);
      if(rem.size() == 0 || (rem.size() == 1 && (*rem.rbegin()).second == x.second))
         return cout << "No" , 0;
      auto it2 = rem.end();
      it2--;
      if((*it2).second == x.second)
         it2--;
      pii y = *it2;
      rem.erase(it2);
      ans[pos[x.second].back()] = y.second;
      pos[x.second].pop_back();
      if(x.first > 1)
         have.insert({x.first - 1 , x.second});
      if(y.first > 1)
         rem.insert({y.first - 1 , y.second});
   }
   cout << "Yes\n";
   for(int i = 0 ; i < n ; i++)
      cout << ans[i] << " ";
   return 0;
}