Artsem has a friend Saunders from University of Chicago. Saunders presented him with the following problem.

Let [n] denote the set {1, ..., n}. We will also write f: [x] → [y] when a function f is defined in integer points 1, ..., x, and all its values are integers from 1 to y.

Now then, you are given a function f: [n] → [n]. Your task is to find a positive integer m, and two functions g: [n] → [m], h: [m] → [n], such that g(h(x)) = x for all <image>, and h(g(x)) = f(x) for all <image>, or determine that finding these is impossible.

Input

The first line contains an integer n (1 ≤ n ≤ 105).

The second line contains n space-separated integers — values f(1), ..., f(n) (1 ≤ f(i) ≤ n).

Output

If there is no answer, print one integer -1.

Otherwise, on the first line print the number m (1 ≤ m ≤ 106). On the second line print n numbers g(1), ..., g(n). On the third line print m numbers h(1), ..., h(m).

If there are several correct answers, you may output any of them. It is guaranteed that if a valid answer exists, then there is an answer satisfying the above restrictions.

Examples

Input

3
1 2 3


Output

3
1 2 3
1 2 3


Input

3
2 2 2


Output

1
1 1 1
2


Input

2
2 1


Output

-1
#include <bits/stdc++.h>
using namespace std;
template <class T>
ostream &operator<<(ostream &os, const vector<T> &vec) {
  os << '{';
  for (int i = (0), __i = (((int)(vec).size())); i < __i; i++) {
    os << vec[i];
    if (i + 1 != ((int)(vec).size())) os << ',';
  }
  os << '}';
  return os;
}
template <class T1, class T2>
ostream &operator<<(ostream &os, const pair<T1, T2> &par) {
  os << '(' << par.first << ',' << par.second << ')';
  return os;
}
int gcd(int x, int y) { return y ? gcd(y, x % y) : abs(x); }
template <class T>
T sqr(T x) {
  return x * x;
}
int f[110 * 1000];
int main() {
  ios::sync_with_stdio(false);
  int n;
  cin >> n;
  vector<int> h(1, 0);
  vector<int> g(n + 1, 0);
  for (int i = (1), __i = (n + 1); i < __i; i++) {
    cin >> f[i];
    if (f[i] == i) {
      g[i] = ((int)(h).size());
      h.push_back(i);
    }
  }
  bool good = true;
  for (int i = (1), __i = (n + 1); i < __i; i++) {
    if (h[g[f[i]]] != f[i]) good = false;
    g[i] = g[f[i]];
  }
  if (!good)
    cout << "-1\n";
  else {
    cout << (((int)(h).size()) - 1) << "\n";
    for (int i = (1), __i = (n + 1); i < __i; i++)
      cout << (" " + (i == 1)) << g[i];
    cout << "\n";
    for (int i = (1), __i = (((int)(h).size())); i < __i; i++)
      cout << (" " + (i == 1)) << h[i];
    cout << "\n";
  }
  return 0;
}
