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)

/*input
1
1

*/
 
/*
                                                    ______________
                                                          |       )
                                                          |   )   /
                                                          |______/
                                                    |     |      \
                                                    |     |   )   \
                                                    |_____|_______)
 
*/
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <cmath>
#include <vector>
#include <set>
#include <map>
#include <unordered_set>
#include <unordered_map>
#include <queue>
#include <ctime>
#include <cassert>
#include <complex>
#include <string>
#include <cstring>
#include <chrono>
#include <random>
#include <queue>
#include <bitset>
#include <iomanip>
#include <fstream>
#include <stack>

using namespace std;
 
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> PII;
typedef pair<ll , ll> PLL;
typedef long double ld;
 
#define pb push_back
#define all(c) c.begin(),c.end()
#define allr(c) c.rbegin(),c.rend()
int mod = 1000000007;
const int inf = 1034567891;
const ll LL_INF = 1234567890123456789ll;
#define PI 3.14159265
#define endl '\n'
#define F first
#define S second
#define debug(x) cout << #x << " = " << x << endl;
#define TRACE
 
#ifdef TRACE
#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
template <typename Arg1>
void __f(const char* name, Arg1&& arg1){
  cout << name << " : " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args){
  const char* comma = strchr(names + 1, ',');cout.write(names, comma - names) << " : " << arg1<<" | ";__f(comma+1, args...);
}
#else
#define trace(...)
#endif
 
#define out(container) for (auto it : container) cout << it << " "; cout << endl;
 
 
template < typename T > T GCD(T a, T b)            { ll t; while(a) { t = a; a = b % a; b = t; } return b; }
template < typename T > string toString(T a)       { return to_string(a); }
template < typename T > void toInt(string s, T &x) { stringstream str(s); str >> x;}
inline int add(int x, int y){ x += y; if(x >= mod) x -= mod; return x;}
inline int sub(int x, int y){ x -= y; if(x < 0) x += mod; return x;}
inline int mul(int x, int y){ return (x * 1ll * y) % mod;}
inline int powr(int a, ll b){
  int x = 1 % mod;
  while(b){
    if(b & 1) x = mul(x, a);
    a = mul(a, a);
    b >>= 1;
  }
  return x;
}
inline int inv(int a){ return powr(a, mod - 2);}

int main()
{
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  
  int n;

  cin >> n;

  vector<int> vec1(n), vec2(n);

  for (auto &it : vec1) {
    cin >> it;
  }  

  for (auto &it : vec2) {
    cin >> it;
  }

  reverse(all(vec2));

  int c = -1;

  int ind = -1;

  for (int i = 0; i < n; i++) {
    if (vec1[i] == vec2[i]) {
      c = vec1[i];
      ind = i;
      break;
    }
  }

  if (c == -1) {
    cout << "Yes" << endl;  
    out(vec2);
    return 0;
  }

  for (int i = 0; i < n; i++) {
    if (ind < n && vec2[ind] == c && vec1[ind] == c && vec1[i] != c) {
      swap(vec1[i], vec2[ind++]);
    }
  }

  if (ind == n || vec2[ind] != vec1[ind]) {
    cout << "Yes" << endl;
    out(vec2);
  } else {
    cout << "No" << endl;
  }


  return 0;
}