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: Time Limit Exceeded)

#include <bits/stdc++.h>
// #include <atcoder/all>
using namespace std;
// using namespace atcoder;
#define rep2(i, m, n) for (int i = (m); i < (n); ++i)
#define rep(i, n) rep2(i, 0, n)
#define drep2(i, m, n) for (int i = (m)-1; i >= (n); --i)
#define drep(i, n) drep2(i, n, 0)
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
#ifdef LOCAL
void debug_out() { cerr << endl; }
template <typename Head, typename... Tail> void debug_out(Head H, Tail... T) { cerr << ' ' << H; debug_out(T...); }
#define debug(...) cerr << 'L' << __LINE__ << " [" << #__VA_ARGS__ << "]:", debug_out(__VA_ARGS__)
#define dump(x) cerr << 'L' << __LINE__ << " " << #x << " = " << (x) << endl
#else
#define debug(...) (void(0))
#define dump(x) (void(0))
#endif
template<typename T> using V = vector<T>;
template<typename T> using P = pair<T, T>;
using ll = long long;
using ld = long double;
using Vi = V<int>;  using VVi = V<Vi>;
using Vl = V<ll>;   using VVl = V<Vl>;
using Vd = V<ld>;   using VVd = V<Vd>;
using Vb = V<bool>; using VVb = V<Vb>;
using Pi = P<int>;
using Pl = P<ll>;
using Pd = P<ld>;
template<typename T> using priority_queue_rev = priority_queue<T, vector<T>, greater<T>>;
template<typename T> vector<T> make_vec(size_t n, T a) { return vector<T>(n, a); }
template<typename... Ts> auto make_vec(size_t n, Ts... ts) { return vector<decltype(make_vec(ts...))>(n, make_vec(ts...)); }
template<typename T> inline int sz(T &x) { return x.size(); }
template<typename T> inline bool chmin(T &a, T b) { if (a > b) { a = b; return true; } return false; }
template<typename T> inline bool chmax(T &a, T b) { if (a < b) { a = b; return true; } return false; }
template<typename T1, typename T2> istream &operator>>(istream &is, pair<T1, T2> &p) { is >> p.first >> p.second; return is; }
template<typename T1, typename T2> ostream &operator<<(ostream &os, const pair<T1, T2> &p) { os << '(' << p.first << ", " << p.second << ')'; return os; }
template<typename T> istream &operator>>(istream &is, vector<T> &v) { for (auto &e : v) is >> e; return is; }
template<typename T> ostream &operator<<(ostream &os, const vector<T> &v) { for (auto &e : v) os << e << ' '; return os; }
template<typename T> inline int count_between(vector<T> &a, T l, T r) { return lower_bound(all(a), r) - lower_bound(all(a), l); } // [l, r)
inline ll cDiv(const ll x, const ll y) { return (x+y-1) / y; } // ceil(x/y)
inline int fLog2(const ll x) { assert(x > 0); return 63-__builtin_clzll(x); } // floor(log2(x))
inline int cLog2(const ll x) { assert(x > 0); return (x == 1) ? 0 : 64-__builtin_clzll(x-1); } // ceil(log2(x))
inline int popcount(const ll x) { return __builtin_popcountll(x); }
inline void fail() { cout << "No" << '\n'; exit(0); }
struct fast_ios { fast_ios(){ cin.tie(nullptr); ios::sync_with_stdio(false); cout << fixed << setprecision(20); }; } fast_ios_;
// const int INF  = 1<<30;
// const ll INFll = 1ll<<60;
// const ld EPS   = 1e-10;
// const ld PI    = acos(-1.0);
// const int MOD  = int(1e9)+7;
// const int MOD  = 998244353;


int main() {
  ll n; cin >> n;

  Vi ca(n), cb(n);
  rep(i, n) {
    int x; cin >> x;
    ++ca[x-1];
  }
  rep(i, n) {
    int x; cin >> x;
    ++cb[x-1];
  }

  rep(i, n) if (ca[i] + cb[i] > n) fail();
  

  set<Pi> Sa, Sb;
  rep(i, n) if (ca[i]) Sa.emplace(ca[i], i);
  rep(i, n) if (cb[i]) Sb.emplace(cb[i], i);



  V<Pi> ab;


  while (!Sa.empty() && !Sa.empty()) {
    auto [c0, i0] = *Sa.rbegin();
    auto [c1, i1] = *Sb.rbegin();

    if (i0 != i1) {
      int k = min(c0, c1);
      rep(i, k) ab.emplace_back(i0, i1);
      Sa.erase({c0, i0});
      if (c0 > k) Sa.emplace(c0 - k, i0);
      Sb.erase({c1, i1});
      if (c1 > k) Sb.emplace(c1 - k, i1);
    }

    else {
      if (sz(Sa) > 1) {
        tie(c0, i0) =  *next(Sa.rbegin());
        int k = min(c0, c1);
        rep(i, k) ab.emplace_back(i0, i1);
        Sa.erase({c0, i0});
        if (c0 > k) Sa.emplace(c0 - k, i0);
        Sb.erase({c1, i1});
        if (c1 > k) Sb.emplace(c1 - k, i1);
      }
      else if (sz(Sb) > 1) {
        tie(c1, i1) =  *next(Sb.rbegin());
        int k = min(c0, c1);
        rep(i, k) ab.emplace_back(i0, i1);
        Sa.erase({c0, i0});
        if (c0 > k) Sa.emplace(c0 - k, i0);
        Sb.erase({c1, i1});
        if (c1 > k) Sb.emplace(c1 - k, i1);
      }
      // else {
      //   fail();
      // }
    }

  }


  cout << "Yes\n";
  sort(all(ab));


  for (auto [x, y] : ab) assert(x != y);

  for (auto [x, y] : ab) cout << (y+1) << ' ';
  cout << '\n';

}