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;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n;
  cin >> n;
  vector<int> F;
  for (int i = 0; i < n; i++) {
    int ele;
    cin >> ele;
    F.push_back(ele);
  }
  vector<int> G = vector<int>(n, -1);
  vector<int> H;
  int cur = 1;
  unordered_set<int> used;
  for (int i = 0; i < n; i++) {
    if (F[i] != F[F[i] - 1]) {
      cout << -1 << endl;
      return 0;
    }
    if (G[i] != -1 || G[F[i] - 1] != -1) {
      G[i] = max(G[i], G[F[i] - 1]);
      G[F[i] - 1] = G[i];
    } else {
      G[i] = cur;
      G[F[i] - 1] = cur;
      cur++;
    }
    if (used.find(F[i]) == used.end()) {
      H.push_back(F[i]);
      used.insert(F[i]);
    }
  }
  cout << H.size() << endl;
  for (auto g : G) {
    cout << g << " ";
  }
  cout << endl;
  for (auto h : H) {
    cout << h << " ";
  }
  cout << endl;
  return 0;
}
