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: Runtime Error)

/*
これを入れて実行
g++ code.cpp
./a.out
 */
 
#include <iostream>
#include <stdio.h>
#include <vector>
#include <string>
#include <cstring>
#include <queue>
#include <deque>
#include <stack>
#include <algorithm>
#include <utility>
#include <set>
#include <map>
#include <unordered_map>
#include <unordered_set>
#include <cmath>
#include <math.h>
#include <tuple>
#include <iomanip>
#include <bitset>
#include <functional>
#include <cassert>
#include <random>

#define all(x) (x).begin(),(x).end()
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
 
using namespace std;
typedef long long ll;
typedef long double ld;
 
int dy4[4] = {-1, 0, +1, 0};
int dx4[4] = {0, +1, 0, -1};
int dy8[8] = {-1, -1, 0, 1, 1, 1, 0, -1};
int dx8[8] = {0, 1, 1, 1, 0, -1, -1, -1};
 
const long long INF = 1LL << 60;
const ll MOD = 1e9 + 7;
 
bool greaterSecond(const pair<int, int>& f, const pair<int, int>& s){
    return f.second > s.second;
}
 
ll gcd(ll a, ll b){
	if (b == 0)return a;
	return gcd(b, a % b);
}
 
ll lcm(ll a, ll b){
    return a / gcd(a, b) * b;
}

ll conbinationMemo[100][100];

void cmemoInit(){
    rep(i, 100){
        rep(j, 100){
            conbinationMemo[i][j] = -1;
        }
    }
}
 
ll nCr(ll n, ll r){
    if(conbinationMemo[n][r] != -1) return conbinationMemo[n][r];
    if(r == 0 || r == n){
        return 1;
    } else if(r == 1){
        return n;
    }
    return conbinationMemo[n][r] = (nCr(n - 1, r) + nCr(n - 1, r - 1));
}
 
ll nPr(ll n, ll r){
    r = n - r;
    ll ret = 1;
    for (ll i = n; i >= r + 1; i--) ret *= i;
    return ret;
}
//-----------------------ここから-----------
int main(void){
    ll n;
    cin >> n;
    vector<ll> a(n), b(n);
    rep(i, n) cin >> a[i];
    rep(i, n) cin >> b[i];;

    vector<ll> rb = b;

    reverse(all(rb));

    vector<ll> same;

    rep(i, n){
        if(a[i] == rb[i]){
            same.push_back(i);
        }
    }

    set<ll> st;
    rep(i, same.size()){
        st.insert(same[i]);
    }

    if(st.size() > 1){
        assert(false);
    }

    vector<ll> ans = rb;
    ll bidx = n - 1;
    rep(i, same.size() / 2){
        swap(ans[same[i]], ans[bidx]);
        bidx--;
    }
    ll fidx = 0;
    for(ll i = (ll)same.size() - 1; i >= (ll)same.size() / 2; i--){
        swap(ans[same[i]], ans[fidx]);
        fidx++;
    }
    bool ok = true;
    rep(i, n){
        if(ans[i] == a[i]) ok = false;
    }

    if(ok){
        cout << "Yes" << endl;
        rep(i, n) cout << ans[i] << " ";
        cout << endl;
    } else {
        cout << "No" << endl;
    }

}