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)

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                 /  \\|||  :  |||//  \
                /  _||||| -:- |||||_  \
                |   | \\\  -  /'| |   |
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          | | :  `- \`. ;`. _/; .'/ /  .' ; |
          \  \ `-.   \_\_`. _.'_/_/  -' _.' /
===========`-.`___`-.__\ \___  /__.-'_.'_.-'================

                  Please give me AC.
*/

#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <numeric>
#include <string>
#include <sstream>
#include <complex>
#include <bitset>
#include <vector>
#include <list>
#include <set>
#include <map>
#include <queue>
#include <deque>
#include <stack>
#include <unordered_map>
#include <unordered_set>
#include <utility>
#include <chrono>
#include <random>

using namespace std;

using int64 = long long;
using uint64 = unsigned long long;
using vi = vector<int>;
using vl = vector<int64>;
using pii = pair<int, int>;
using pll = pair<int64, int64>;

#define rep(i, n) for (int i = 0; i < (int)(n); ++i)
#define all(v) (v).begin(), (v).end()
#define print(x) cout << (x) << '\n'
#define print2(x, y) cout << (x) << ' ' << (y) << '\n'
#define print3(x, y, z) cout << (x) << ' ' << (y) << ' ' << (z) << '\n'
#define printn(v) rep(i, (v).size() - 1) cout << (v)[i] << ' '; cout << (v)[n - 1] << '\n';
#ifdef ONLINE_JUDGE
#define debug(x)
#define debug2(x, y)
#define debug3(x, y, z)
#define dbg(v)
#else
#define debug(x) cerr << #x << ": " << (x) << '\n'
#define debug2(x, y) cerr << #x << ": " << (x) << ", " << #y << ": " << (y) << '\n'
#define debug3(x, y, z) cerr << #x << ": " << (x) << ", " << #y << ": " << (y) << ", " << #z << ": " << (z) << '\n'
#define dbg(v) for (size_t _ = 0; _ < v.size(); ++_){cerr << #v << "[" << _ << "] : " << v[_] << '\n';}
#endif

// constant
const int INF = (1<<30) - 1;
const int64 INF64 = (1LL<<62) - 1;

template<typename T> T gcd(T a, T b) {
    if (a < b) return gcd(b, a);
    T r;
    while ((r = a % b)) {
        a = b;
        b = r;
    }
    return b;
}

template<typename T> T lcm(const T a, const T b) {
    return a / gcd(a, b) * b;
}

template<typename T> bool chmin(T& a, const T& b) {
    if (a > b) return a = b, true; else return false;
}

template<typename T> bool chmax(T& a, const T& b) {
    if (a < b) return a = b, true; else return false;
}

// End of template.

int n, a[200200], b[200200], ca[200200], cb[200200], ans[200200];


int main() {
    cout << fixed << setprecision(15);
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);
    cerr.tie(nullptr);

    cin >> n;
    rep(i, n) {
        cin >> a[i];
        ca[a[i]] += 1;
    }
    rep(i, n) {
        cin >> b[i];
        cb[b[i]] += 1;
    }

    for (int i = 1; i <= n; ++i) {
        if (ca[i] + cb[i] > n) {
            print("No");
            return 0;
        }
    }

    rep(i, n) {
        ca[i + 1] += ca[i];
        cb[i + 1] += cb[i];
    }

    int s = 0;
    for (int i = 1; i <= n; ++i) {
        chmax(s, cb[i] - ca[i - 1]);
    }

    rep(i, n) {
        ans[i] = b[(i + s) % n];
    }

    print("Yes");
    rep(i, n - 1) cout << ans[i] << ' ';
    cout << ans[n - 1] << '\n';

    return 0;
}

/*
  __
 / _| ___ _ __  _ __
| |_ / __| '_ \| '_ \
|  _| (__| |_) | |_) |
|_|(_)___| .__/| .__/
         |_|   |_|
*/