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)

#include <iostream>
#include <algorithm>
#include <utility>
#include <vector>
#include <string>
#include <map>

using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
#define REP(i, n) for (int i = 0; i < (n); i++)
#define RREP(i, n) for (int i = (n) - 1; i >= 0; i--)
#define FOR(i, a, b) for (int i = (a); i < (b); i++)
#define RFOR(i, a, b) for (int i = (a); i > (b); i--)
#define ALL(a) (a).begin(), (a).end()

int main() {
    int N, A[200000], B[200000];
    cin >> N;
    map<int, int> mii;
    REP(i, N) {
        cin >> A[i];
        mii[A[i]]++;
    }
    REP(i, N) {
        cin >> B[i];
        mii[B[i]]++;
    }
    for (auto p : mii) {
        if (p.second > N) {
            cout << "No" << endl;
            return 0;
        }
    }

    int C[200001] = {}, D[200001] = {};
    REP(i, N) {
        C[A[i]]++;
        D[B[i]]++;
    }
    REP(i, N) {
        C[i + 1] += C[i];
        D[i + 1] += D[i];
    }

    int offset = 0;
    REP(i, N) {
        offset = max(D[i + 1] - C[i], offset);
    }
    cout << "Yes" << endl;
    REP(i, N) {
        cout << B[(i + offset) % N] << " ";
    }
    cout << endl;
}