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: Time Limit Exceeded)

#include <bits/stdc++.h>

using namespace std;

typedef long long int ll;
typedef pair<ll, ll> pll;

#define FOR(i, n, m) for(ll (i)=(m);(i)<(n);++(i))
#define REP(i, n) FOR(i,n,0)
#define OF64 std::setprecision(10)

const ll MOD = 1000000007;
const ll INF = (ll) 1e15;

ll A[200005];
ll B[200005];
pll NA[200005];
pll NB[200005];

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);

    ll N;
    cin >> N;
    memset(NA, 0, sizeof(NA));
    memset(NB, 0, sizeof(NB));
    REP(i, N) {
        cin >> A[i];
        NA[A[i] - 1].first = A[i];
        NA[A[i] - 1].second++;
    }
    REP(i, N) {
        cin >> B[i];
        NA[B[i] - 1].first = B[i];
        NA[B[i] - 1].second++;
    }

    REP(i, 200005) {
        ll mx = std::max(NA[i].second, NB[i].second);
        ll mn = std::min(NA[i].second, NB[i].second);
        if (N - mx < mn) {
            cout << "No" << endl;
            return 0;
        }
    }
    cout << "Yes" << endl;

    queue<ll> q;
    REP(i, N) {
        q.push(B[i]);
    }

    REP(i, N) {
        while (true) {
            ll t = q.front();
            q.pop();
            if (t != A[i]) {
                B[i] = t;
                break;
            }
            q.push(t);
        }
    }

    REP(i, N) {
        cout << B[i] << " ";
    }
    cout << endl;
    return 0;
}