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;
#define int long long
#define ld long double
#define ll long long
#define P pair<int,int>
#define fi first
#define se second
#define rep(i,n) for(int i=0;i<n;i++)
#define all(v) v.begin(),v.end()
#define pb push_back
#define eb emplace_back
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;}
constexpr int INF=1000000000000000000;
constexpr int mod=1000000007;
constexpr double eps=1e-8;
const double pi=acos(-1);
int dx[]={0,1,0,-1},dy[]={1,0,-1,0};
int Random(int mi,int ma){
    random_device rnd;
    mt19937 mt(rnd());//32bit
    //[mi,ma]
    uniform_int_distribution<int>engine(mi,ma);
    return engine(mt);
}
int gcd(int a,int b){
    if(b==0)return a;
    return gcd(b,a%b);
}
int lcm(int a,int b){
    return a/gcd(a,b)*b;
}
bool prime(int a){
    if(a==1)return false;
    for(int i=2;i*i<=a;i++){
        if(a%i==0)return false;
    }
    return true;
}
int modpow(int a,int b){
    if(b==0)return 1;
    if(b%2)return modpow(a,b-1)*a%mod;
    int memo=modpow(a,b/2);
    return memo*memo%mod;
}
vector<int>kaijo,invkaijo;
void init_fact(int n){
    kaijo.resize(n+1);
    invkaijo.resize(n+1);
    kaijo[0]=1;
    for(int i=1;i<=n;i++){
        kaijo[i]=kaijo[i-1]*i;
        kaijo[i]%=mod;
    }
    rep(i,n+1)invkaijo[i]=modpow(kaijo[i],mod-2);
}
int comb(int a,int b){
    if(a<b)return 0;
    if(a<0||b<0)return 0;
    return kaijo[a]*invkaijo[a-b]%mod*invkaijo[b]%mod;
}
int inv(int x){
    x=modpow(x,mod-2);
    return x;
}
int n,a[200010],b[200010],ans[200010];
set<int>st;
signed main(){
    cin.tie(0);ios::sync_with_stdio(false);
    cin>>n;
    rep(i,n)cin>>a[i];
    rep(i,n)cin>>b[i];
    rep(i,n)st.insert(i);
    rep(i,n){
        auto it=st.end();
        it--;
        if(a[*it]!=b[i]){
            ans[*it]=b[i];
            st.erase(it);
        }
        else{
            int x=lower_bound(a,a+n,*it)-a;
            it=st.lower_bound(x);
            if(it==st.begin()){
                puts("No");
                return 0;
            }
            it--;
            ans[*it]=b[i];
            st.erase(it);
        }
    }
    cout<<"Yes"<<endl;
    rep(i,n)cout<<ans[i]<<" ";
    cout<<endl;
    return 0;
}