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 vector<int> vi;
typedef vector<ll> vll;
typedef vector<vi> vvi;
typedef vector<string> vs;
typedef pair<int, int> pii;

#define sz(a) int((a).size())
#define pb push_back
#define eb emplace_back
#define fi first
#define se second
#define all(c) (c).begin(),(c).end()
#define rall(c) (c).rbegin(),(c).rend()
#define ini(a, i) memset(a, i, sizeof(a))
#define inin(a, n, i) memset(a, i, sizeof(a[0])*(n))
#define contains(c, i) ((c).find(i) != (c).end())
#define present(i, c) (find(all(c), i) != (c).end())

#define trav(x, c) for(auto& x : c)
#define rep(i, n) for(int i = 0; i < n; i++)
#define repa(i, b, e) for(int i = (b)-((b)>(e)); i != (e)-((b)>(e)); i += 1-2*((b)>(e)))
	
template<class T> bool chkmax(T &x, T y) { return x < y ? x = y, true : false; }
template<class T> bool chkmin(T &x, T y) { return x > y ? x = y, true : false; }
template<class A, class B> A cvt(B x) { stringstream s; s << x; A r; s >> r; return r; }

void read() {}
void print() {}
template<class T, class... Args> void read(T& a, Args&... args) { cin >> a; read(args...); }
template<class T, class... Args> void print(T a, Args... args) { cout << a << ' '; print(args...); }
template<class... Args> void println(Args... args) { print(args...); cout << '\n'; }

#define debug(args...) { string s_(#args); replace(all(s_), ',', ' '); stringstream ss_(s_); istream_iterator<string> it_(ss_); cerr_(it_, args); }
void cerr_(istream_iterator<string> it) { (void) it; }
template<class T, class... Args> void cerr_(istream_iterator<string> it, T a, Args... args) { cerr << *it << " = " << a << endl; cerr_(++it, args...); }

template<class T, class S>  ostream& operator<<(ostream& os, const pair<T, S>& p) { os << "(" << p.first << ", " << p.second << ")"; return os; }
template<class T> ostream& operator<<(ostream &os, const vector<T> &v) { os << '{'; string sep; for (const auto &x : v) os << sep << x, sep = ", "; return os << '}'; }
template<class T> ostream& operator<<(ostream &os, const set<T> &s) { os << '{'; string sep; for (const auto &x : s) os << sep << x, sep = ", "; return os << '}'; }
template<class T, class S> ostream& operator<<(ostream &os, const map<T, S> &m) { os << '{'; string sep; for (const auto &x : m) os << sep << x, sep = ", "; return os << '}'; }
template<class T, size_t size> ostream& operator<<(ostream &os, const array<T, size> &arr) { os << '{'; string sep; for (const auto &x : arr) os << sep << x, sep = ", "; return os << '}'; }

const int INF = 0x3F3F3F3F;
const int MAXN = (int) 1e6;
const int MOD = (int) 1e9+7;

//=========================

vi solve(vi &a, vi &b) {
    deque<int> d(all(b));
    vi ans;
    rep(i, sz(a)) {
        bool ok = false;
        if(d.back() != a[i]) {
            ans.pb(d.back());
            d.pop_back();
            ok = true;
        }
        if(!ok && d.front() != a[i]) {
            ans.pb(d.front());
            d.pop_front();
            ok = true;
        }
        if(!ok) {
            return {};
        }
    }
    return ans;
}

void run_test() {
    int n; read(n);
    vi a(n), b(n);
    rep(i, n) read(a[i]);
    rep(i, n) read(b[i]);

    vi ans = solve(a, b);
    if(sz(ans) == 0) {
        reverse(all(b));
        ans = solve(a, b);
    }

    if(sz(ans) == 0) {
        println("No");
        return;
    }

    println("Yes");
    trav(x, ans) print(x);
    println();
}

//=========================

int main() {
	ios::sync_with_stdio(false); cin.tie(nullptr);
	//int t_ = 1, t__; cin >> t__; while(t_ <= t__) { cout << "Case #" << t_++ << ": "; run_test(); }
	// int t_; cin >> t_; while(t_--) run_test();
	run_test();
	return 0;	
}