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 <typename T>
ostream& operator<<(ostream& o, vector<T>& v) {
  for (auto& x : v) o << x << ' ';
  return o;
}
int main() {
  std::ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  vector<int> arr(n);
  for (int i = 0; i < n; i++) cin >> arr[i];
  vector<int> idem(n, 0);
  set<int> S;
  for (int i = 0; i < n; i++) {
    if (arr[i] == i + 1) idem[i] = 1;
    S.insert(arr[i]);
  }
  for (int x : S) {
    if (!idem[x - 1]) {
      cout << -1 << endl;
      return 0;
    }
  }
  vector<int> g, h;
  vector<int> seen(n + 1, -1);
  int cur = 1;
  for (int i = 0; i < n; i++) {
    if (seen[arr[i]] == -1) {
      h.push_back(arr[i]);
      seen[arr[i]] = cur++;
    }
    g.push_back(seen[arr[i]]);
  }
  cout << (int)h.size() << endl;
  cout << g << endl;
  cout << h << endl;
}
