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;
const int INF = 0x3f3f3f3f;
const LL mod = 1e9 + 7;
const int N = 200005;
int a[N], b[N], c[N];
int main() {
    int n;
    cin >> n;
    for (int i = 0; i < n; i++) cin >> a[i], c[a[i]]++;
    for (int i = 0; i < n; i++) cin >> b[i], c[a[i]]++;
    for (int i = 1; i <= n; i++) {
        if (c[i] > n) {
            return puts("No"), 0;
        }
    }
    reverse(b, b + n);
    int pos = 0;
    for (int i = 0; i < n; i++) {
        if (a[i] == b[i]) {
            while (a[pos] == a[i] || b[pos] == a[i]) pos++;
            swap(b[pos],b[i]);
            pos++;
        }
    }
    puts("Yes");
    for (int i = 0; i < n; i++) printf("%d%c", b[i], " \n"[i + 1 == n]);
    return 0;
}