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 <bits/stdc++.h>

using namespace std;
typedef long long int ll;
typedef pair<int, int> ii;
#define DEBUG freopen("in.txt", "r", stdin);

struct fastio {
    fastio() {
        ios::sync_with_stdio(false);
        cout << setprecision(10) << fixed;
        cin.tie(0);
    }
};

fastio _fast_io;

const int N = 2e5 + 5;
int n, t;
int a[N], b[N], ACnt[N], BCnt[N];

int main() {
    cin >> n;
    for (int i = 0; i < n; ++i) {
        cin >> t;
        a[i] = t;
        ++ACnt[t];
    }
    for (int i = 0; i < n; ++i) {
        cin >> t;
        b[i] = t;
        ++BCnt[t];
    }
    for (int i = 0; i < N; ++i) {
        if (ACnt[i] + BCnt[i] > n) {
            cout << "No" << endl;
            exit(0);
        }
    }
    reverse(b, b + n);
    int left = 0, right = n - 1;
    for (int i = 0; i < n; ++i) {
        if (b[i] == a[i]) {
            if (i - left >= right - i) {
                swap(b[i], b[left]);
                left++;
            } else {
                swap(b[i], b[right]);
                --right;
            }
        }
    }
    cout << "Yes" << endl;
    for (int i = 0; i < n; ++i) {
        cout << b[i] << ' ';
    }
    cout << endl;

    return 0;
}