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;
double PI = 4.00 * (atan(1));
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long int n;
  cin >> n;
  long long int fun[n + 1];
  set<long long int> s1;
  for (long long int i = 1; i <= n; i++) {
    cin >> fun[i];
  }
  bool check = true;
  for (long long int i = 1; i <= n; i++) {
    if (fun[fun[i]] != fun[i]) {
      check = false;
    } else {
      s1.insert(fun[i]);
    }
  }
  if (check == false) {
    cout << "-1";
    exit(0);
  }
  vector<long long int> final(s1.begin(), s1.end());
  long long int size = final.size();
  map<long long int, long long int> pos;
  for (long long int i = 0; i < size; i++) {
    pos[final[i]] = i + 1;
  }
  cout << size << endl;
  for (long long int i = 1; i <= n; i++) {
    cout << pos[fun[i]] << " ";
  }
  cout << endl;
  for (long long int i = 0; i < size; i++) {
    cout << final[i] << " ";
  }
  return 0;
}
