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: Accepted)

#include <bits/stdc++.h>

/*
#include <atcoder/all>
using namespace atcoder;
*/
#pragma GCC target("avx2")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
//#include <boost/multiprecision/cpp_int.hpp>
using namespace std;


using dou =long double;
string yes="yes";
string Yes="Yes";
string YES="YES";
string no="no";
string No="No";
string NO="NO";

template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }
typedef long long ll;
typedef pair<int,int> P;
typedef pair<ll,ll> PL;
const ll mod = 1000000007ll;
//const ll mod = 10000000000ll;
//const ll mod = 10000;

struct mint {
  ll x; // typedef long long ll;
  mint(ll x=0):x((x%mod+mod)%mod){}
  mint operator-() const { return mint(-x);}
  mint& operator+=(const mint a) {
    if ((x += a.x) >= mod) x -= mod;
    return *this;
  }
  mint& operator-=(const mint a) {
    if ((x += mod-a.x) >= mod) x -= mod;
    return *this;
  }
  mint& operator*=(const mint a) { (x *= a.x) %= mod; return *this;}
  mint operator+(const mint a) const { return mint(*this) += a;}
  mint operator-(const mint a) const { return mint(*this) -= a;}
  mint operator*(const mint a) const { return mint(*this) *= a;}
  mint pow(ll t) const {
    if (!t) return 1;
    mint a = pow(t>>1);
    a *= a;
    if (t&1) a *= *this;
    return a;
  }

  // for prime mod
  mint inv() const { return pow(mod-2);}
  mint& operator/=(const mint a) { return *this *= a.inv();}
  mint operator/(const mint a) const { return mint(*this) /= a;}
};
istream& operator>>(istream& is, const mint& a) { return is >> a.x;}
ostream& operator<<(ostream& os, const mint& a) { return os << a.x;}
#define rep(i, n)         for(ll i = 0; i < (ll)(n); i++)
#define brep(n)           for(int bit=0;bit<(1<<n);bit++)
#define bbrep(n)           for(int bbit=0;bbit<(1<<n);bbit++)
#define erep(i,container) for (auto &i : container)
#define itrep(i,container) for (auto i : container)
#define irep(i, n)        for(ll i = n-1; i >= (ll)0ll; i--)
#define rrep(i,m,n) for(ll i = m; i < (ll)(n); i++)
#define reprep(i,j,h,w) rep(i,h)rep(j,w)
#define all(x) (x).begin(),(x).end()
#define rall(x) (x).rbegin(),(x).rend()
#define VEC(type,name,n) std::vector<type> name(n);rep(i,n)std::cin >> name[i];
#define pb push_back
#define pf push_front
#define query int qq;std::cin >> qq;rep(qqq,qq)
#define lb lower_bound
#define ub upper_bound
#define fi first
#define se second
#define itn int
#define mp make_pair
//#define sum(a) accumulate(all(a),0ll)
#define keta fixed<<setprecision
#define vout(a) erep(qxqxqx,a)std::cout << qxqxqx << ' ';std::cout  << std::endl;
#define vvector(name,typ,m,n,a)vector<vector<typ> > name(m,vector<typ> (n,a))
//#define vvector(name,typ,m,n)vector<vector<typ> > name(m,vector<typ> (n))
#define vvvector(name,t,l,m,n,a) vector<vector<vector<t> > > name(l, vector<vector<t> >(m, vector<t>(n,a)));
#define vvvvector(name,t,k,l,m,n,a) vector<vector<vector<vector<t> > > > name(k,vector<vector<vector<t> > >(l, vector<vector<t> >(m, vector<t>(n,a)) ));
#define case std::cout <<"Case #" <<qqq+1<<":"
#define RES(a,i,j) a.resize(i);rep(ii,i)a[ii].resize(j);
#define RESRES(a,i,j,k) a.resize(i);rep(ii,i)a[ii].resize(j);reprep(ii,jj,i,j){dp[ii][jj].resize(k)};
#define res resize
#define as assign
#define ffor for(;;)
#define ppri(a,b) std::cout << a<<" "<<b << std::endl
#define pppri(a,b,c) std::cout << a<<" "<<b <<" "<< c<<std::endl
#define ppppri(a,b,c,d) std::cout << a<<" "<<b <<" "<< c<<' '<<d<<std::endl
#define aall(x,n) (x).begin(),(x).begin()+(n)
#define SUM(a) accumulate(all(a),0ll) 
#define stirng string
#define gin(a,b) int a,b;std::cin >> a>>b;a--;b--;
#define popcount __builtin_popcount
#define permu(a) next_permutation(all(a))
//#define grid_input(a,type) int h,w;std::cin >> h>>w;vvector(a,type,h,w,0);reprep(i,j,h,w)std::cin >> a[i][j];

//typedef long long T;
ll ceil(ll a,ll b){
    return ((a+b-1)/b);
}
const int INF = 2000000000;
//const ll INF64 =3223372036854775807ll;
//const ll INF64 = 9223372036854775807ll;
const ll INF64 = 243000000000001ll;;;
const ll MOD = 1000000007ll;
//const ll MOD = 1000003ll;
const ll OD = 1000000000000007ll;
const dou pi=3.141592653589793;
long long modpow(long long a, long long n) { //累乗の余剰
    long long res = 1;
    while (n > 0) {
        if (n & 1) res = res * a % MOD;
        a = a * a % MOD;
        n >>= 1;
    }
    return res;
}


//メモ
//ゲーム(Grundy数とか)の復習をする
//群論の勉強をする？
//ドツボにハマったら頑張って今までの思考をリセットする
//学会のスライドを作る(水曜日まで)
//周期性の実験をする
//リスニング力をどうにかする
//GT学会の会費を払う
/*
#include <atcoder/all>
using namespace atcoder;
*/
/*
#include <atcoder/all>
using namespace atcoder;
*/
int main(){
    int n;
    std::cin >> n;
    VEC(int,a,n);
    VEC(int,b,n);
    std::vector<int> c(n+1),cc(n+1);
    rep(i,n){
        c[a[i]]++;
        cc[b[i]]++;
    }
    rep(i,n+1){
        if(c[i]>n-cc[i]||cc[i]>n-c[i]){
            std::cout << No << std::endl;
            exit(0);
        }
    }
    sort(rall(b));
    std::priority_queue<
      int,                // 要素の型はint
      std::vector<int>,   // 内部コンテナはstd::vector (デフォルトのまま)
      std::greater<int>   // 昇順 (デフォルトはstd::less<T>)
    > ichi,le;
    priority_queue<int>ri;
    bool owa=0;
    rep(i,n){
        if(a[i]==b[i]){
            ichi.push(i);
            owa=1;
        }
        else {
            if(owa)ri.push(i);
            else le.push(i);
        }
    }
    //pppri(ichi.size(),le.size(),ri.size());
   // vout(a);
    //vout(b);
    while(!ichi.empty()){
    //    pppri(ichi.size(),le.size(),ri.size());
        if(ri.size()!=0){
            
            int p=ichi.top();
            int pp=ri.top();
           // ppri(p,pp);
            if(b[p]==a[pp]||b[p]==b[pp]){
                while(ri.size()!=0){
                    ri.pop();
                }
                goto to;
            }
            swap(b[p],b[pp]);
            ichi.pop();
            ri.pop();
        }
        else{
to:         int p=ichi.top();
            int pp=le.top();
            swap(b[p],b[pp]);
            ichi.pop();
            le.pop();
        }
    }
    std::cout << Yes << std::endl;
    vout(b);
}