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)

// 2020-09-14 20:57:26
// clang-format off
#include<bits/stdc++.h>
#ifdef LOCAL
#include "lib/debug.hpp"
#else
#define debug(...) 1
#endif
#define ALL(a) (a).begin(), (a).end()
#define RALL(a) (a).rbegin(), (a).rend()
#define rep(i, n) REP(i, 0, (n))
#define repc(i, n) REPC(i, 0, (n))
#define REP(i, n, m) for (int i = (int)(n); i < (int)(m); i++)
#define REPC(i, n, m) for (int i = (int)(n); i <= (int)(m); i++)
#define REPCM(i, n, m) for (int i = (int)(n); i >= (int)(m); i--)
using namespace std;
using ll = long long;
using ld = long double;
using pr = pair<ll, ll>;
using vll = vector<ll>;
using vpr = vector<pr>;
using P = pair<int, int>;
template<class T> inline bool chmin(T& a, const T& b) { if (a > b) { a = b; return true; } else return false; }
template<class T> inline bool chmax(T& a, const T& b) { if (a < b) { a = b; return true; } else return false; }

// clang-format on
void answer() {
  int n;
  cin >> n;
  vector<int> a(n), b(n);
  rep(i, n) cin >> a[i];
  rep(i, n) cin >> b[i];
  int d = 0;
  rep(ai, n) {
    auto itr = lower_bound(ALL(b), a[ai]);
    if (itr != b.end() && a[ai] == *itr) {
      chmax(d, ai - int(itr - b.begin() - 1));
    }
  }
  bool ok = true;
  rep(i, n) ok &= (a[i] != b[(n + i - d) % n]);
  if (ok) {
    cout << "Yes" << '\n';
    rep(i, n) {
      if (i) cout << ' ';
      cout << b[(n + i - d) % n];
    }
    cout << '\n';
  } else {
    cout << "No" << '\n';
  }
}
int main() {
  std::ios::sync_with_stdio(false);
  std::cin.tie(0);
  answer();
  return 0;
}