#define __USE_MINGW_ANSI_STDIO 0
#include <bits/stdc++.h>

using namespace std;
using ll = long long;
#define int ll
using VI = vector<int>;
using VVI = vector<VI>;
using PII = pair<int, int>;

#define FOR(i, a, n) for (ll i = (ll)a; i < (ll)n; ++i)
#define REP(i, n) FOR(i, 0, n)
#define ALL(x) x.begin(), x.end()
#define PB push_back

const ll LLINF = (1LL<<60);
const int INF = (1LL<<30);
const int MOD = 1000000007;

template <typename T> T &chmin(T &a, const T &b) { return a = min(a, b); }
template <typename T> T &chmax(T &a, const T &b) { return a = max(a, b); }
template <typename T> bool IN(T a, T b, T x) { return a<=x&&x<b; }
template<typename T> T ceil(T a, T b) { return a/b + !!(a%b); }
template<class S,class T>
ostream &operator <<(ostream& out,const pair<S,T>& a){
  out<<'('<<a.first<<','<<a.second<<')';
  return out;
}
template<class T>
ostream &operator <<(ostream& out,const vector<T>& a){
  out<<'[';
  REP(i, a.size()) {out<<a[i];if(i!=a.size()-1)out<<',';}
  out<<']';
  return out;
}

int dx[] = {0, 1, 0, -1}, dy[] = {1, 0, -1, 0};

signed main(void)
{
  cin.tie(0);
  ios::sync_with_stdio(false);

  int n;
  cin >> n;
  VI a(n), b(n);
  REP(i, n) cin >> a[i];
  REP(i, n) cin >> b[i];

  set<int> st;
  REP(i, n) st.insert(a[i]);
  REP(i, n) {
    if(st.find(b[i]) != st.end()) {
      st.erase(b[i]);
    }
  }

  if(st.size() == 0) {
    cout << "Yes" << endl;
    return 0;
  }

  // stに残っているところを not equal にする
  VI c(n), d(n);
  REP(i, n) {
    if(st.find(a[i]) == st.end()) {
      c[i] = a[i], d[i] = a[i];
    } else {
      c[i] = a[0], d[i] = a[1];
    }
  }

  cout << "No" << endl;
  REP(i, n) cout << c[i] << (i==n-1?"\n":" ");
  REP(i, n) cout << d[i] << (i==n-1?"\n":" ");

  return 0;
}


### Prompt

Develop a solution in CPP to the problem described below:

problem

AOR Ika made a set $ S = \\ {a_1, ..., a_N \\} $ and a map $ f: S → S $. $ f (a_i) = b_i $. For any element $ x $ in the set $ S $, all maps $ g, h: S → S $ satisfying $ g (f (x)) = h (f (x)) $ are $ g (x). ) = Determine if h (x) $ is satisfied, and if not, configure one counterexample.





Example

Input

5
1 2 3 4 5
3 4 2 5 1


Output

Yes

### Response

```cpp
#define __USE_MINGW_ANSI_STDIO 0
#include <bits/stdc++.h>

using namespace std;
using ll = long long;
#define int ll
using VI = vector<int>;
using VVI = vector<VI>;
using PII = pair<int, int>;

#define FOR(i, a, n) for (ll i = (ll)a; i < (ll)n; ++i)
#define REP(i, n) FOR(i, 0, n)
#define ALL(x) x.begin(), x.end()
#define PB push_back

const ll LLINF = (1LL<<60);
const int INF = (1LL<<30);
const int MOD = 1000000007;

template <typename T> T &chmin(T &a, const T &b) { return a = min(a, b); }
template <typename T> T &chmax(T &a, const T &b) { return a = max(a, b); }
template <typename T> bool IN(T a, T b, T x) { return a<=x&&x<b; }
template<typename T> T ceil(T a, T b) { return a/b + !!(a%b); }
template<class S,class T>
ostream &operator <<(ostream& out,const pair<S,T>& a){
  out<<'('<<a.first<<','<<a.second<<')';
  return out;
}
template<class T>
ostream &operator <<(ostream& out,const vector<T>& a){
  out<<'[';
  REP(i, a.size()) {out<<a[i];if(i!=a.size()-1)out<<',';}
  out<<']';
  return out;
}

int dx[] = {0, 1, 0, -1}, dy[] = {1, 0, -1, 0};

signed main(void)
{
  cin.tie(0);
  ios::sync_with_stdio(false);

  int n;
  cin >> n;
  VI a(n), b(n);
  REP(i, n) cin >> a[i];
  REP(i, n) cin >> b[i];

  set<int> st;
  REP(i, n) st.insert(a[i]);
  REP(i, n) {
    if(st.find(b[i]) != st.end()) {
      st.erase(b[i]);
    }
  }

  if(st.size() == 0) {
    cout << "Yes" << endl;
    return 0;
  }

  // stに残っているところを not equal にする
  VI c(n), d(n);
  REP(i, n) {
    if(st.find(a[i]) == st.end()) {
      c[i] = a[i], d[i] = a[i];
    } else {
      c[i] = a[0], d[i] = a[1];
    }
  }

  cout << "No" << endl;
  REP(i, n) cout << c[i] << (i==n-1?"\n":" ");
  REP(i, n) cout << d[i] << (i==n-1?"\n":" ");

  return 0;
}


```