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 long double ld;
#define REP(i,n) for(int i=0, i##_len=(n); i<i##_len; ++i)
#define REPR(i,n) for(int i=n; i>-1; --i)
#define ALL(a) (a).begin(),(a).end()
#define FILL(a,n,x); REP(i,(n)){ (a)[i]=(x); }
#define CINA(a,n); REP(i,(n)){ cin >> (a)[i]; }
#define FILL2(a,n,m,x); REP(i,(n)){ REP(j,(m)){(a)[i][j]=(x);} }
#define CINA2(a,n,m); REP(i,(n)){ REP(j,(m)){cin >> (a)[i][j];} }
#define Liny "Yes\n"
#define Linn "No\n"
#define LINY "YES\n"
#define LINN "NO\n"

//cout << setfill('0') << right << setw`(4) << 12; // "0012"
int keta(ll x){ if(x<10){return 1;} else{return keta(x/10) + 1;}}
int keta_wa(ll x){ if(x<10){return x;} else{return keta_wa(x/10) + x%10;} }
int ctoi(char c){ return ( (c>='0' && c<='9')? c - '0': 0 );}
int __stoi(string s){ return atoi(s.c_str()); }
ll sum(ll a[],ll N){ return accumulate(a,a+N,0LL);}
ll gcd(ll a,ll b){if(a<b)swap(a,b); return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b){if(a<b){swap(a,b);} return a/gcd(a,b)*b;}
template<class T> void chmax(T& a, T b){ if(a<b){a=b;} }
template<class T> void chmin(T& a, T b){ if(a>b){a=b;} }
template<class T> bool isIn(T a,vector<T> v){ for(T x:v){ if(a==x){return true;}} return false;}
string strReplace(string s,string target, string replacement){
    if (!target.empty()) {
      std::string::size_type pos = 0;
      while ((pos = s.find(target, pos)) != std::string::npos) {
        s.replace(pos, target.length(), replacement);
        pos += replacement.length();
      }
    }
    return s;
}
const ll MOD = 1e9+7;
#define pii pair<int,int>
#define pll pair<ll,ll>
#define MP make_pair

bool DEBUG = 0;
template<class T> void dprint(T s){
  if(DEBUG){ cout << s << "\n"; }
}


ll a[200005],b[200005];
map<ll,ll> mpa,mpb;

void solve(){
  ll N; cin>>N;
  REP(i,N){
    cin >> a[i];
    ++mpa[ a[i] ];
  }
  REP(i,N){
    cin >> b[i];
    ++mpb[ b[i] ];
  }
  REP(x,N+1){
    if(mpa[x]+mpb[x]>N){
      cout << "No\n";
      return;
    }
  }
  sort(b,b+N,greater<ll>());
  int idx = 0;
  int temp = -1;
  REP(i,N){
    if(a[i]==b[i]){
      if(temp!=a[i]){
        idx=0;
      }
      while(a[i]==b[idx] && idx<N){
        ++idx;
      }
      if(idx==N){cout << Linn; return;}
      swap(b[idx],b[i]);
    }
    temp = a[i];
  }
  cout << Liny;
  REP(i,N){
    if(i<N-1){
      printf("%Ld ",b[i]);
    }else{
      printf("%Ld\n",b[i]);
    }
  }
}
int main(int argc, char *argv[]){
  if(argc>1 && argv[1][0]=='D'){ DEBUG = 1;}
  solve();
  return 0;
}