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: Runtime Error)

#include<bits/stdc++.h>

using namespace std;

typedef long long int ll;

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

  int n;
  cin >> n;
  vector < int > a (n), b (n);
  for (auto &i : a) cin >> i;
  for (auto &i : b) cin >> i;

  reverse (b.begin(), b.end());
  int l = -1, r = -2;
  for (int i = 0; i < n; i++) {
    if (a[i] == b[i]) {
      if (l == -1) l = r = i;
      else r++;
    }
  }

  if (l == -1) {
    cout << "Yes\n";
    for (auto i : b) cout << i << " ";
    cout << '\n';
    return 0;
  }

  int p = lower_bound (a.begin(), a.end(), a[l]) - a.begin() - 1;
  int q = upper_bound (a.begin(), a.end(), a[l]) - a.begin();
  q = max (q, r + 1);
  p = min (p, l - 1);
  int idx = 0;
  if (p <= 0) idx = q + 1;
  for (int i = l; i <= r; i++) {
    if (idx == n) {
      cout << "No\n";
      return 0;
    }
    swap (b[i], b[idx]);
    idx++;
    if (idx == p) idx = q + 1;
  }

  cout << "Yes\n";
  for (auto i : b) cout << i << " ";
  cout << '\n';
}