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;
ostream& operator<<(ostream& out, pair<int64_t, int64_t>& p) {
  out << "{" << p.first << ", " << p.second << "} ";
  return out;
}
template <typename T>
istream& operator>>(istream& in, vector<T>& arr) {
  for (auto& i : arr) in >> i;
  return in;
}
template <typename T>
ostream& operator<<(ostream& out, vector<T>& arr) {
  for (auto i : arr) out << i << " ";
  out << '\n';
  return out;
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int64_t n;
  cin >> n;
  vector<int64_t> f(n);
  cin >> f;
  int64_t m = 0;
  vector<int64_t> g(n, -1);
  vector<int64_t> h;
  for (int64_t i = 0; i < n; ++i) {
    if (f[i] == i + 1) {
      ++m;
      g[i] = m;
      h.push_back(i + 1);
    }
  }
  for (int64_t i = 0; i < n; ++i) {
    if (g[f[i] - 1] == -1) {
      cout << -1 << '\n';
      return 0;
    } else if (f[i] != f[f[i] - 1]) {
      cout << -1 << '\n';
      return 0;
    } else {
      g[i] = g[f[i] - 1];
    }
  }
  if (m == 0) {
    cout << -1 << '\n';
    return 0;
  }
  cout << m << '\n';
  cout << g;
  cout << h << '\n';
  return 0;
}
