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);
  int n;
  cin >> n;
  int i, j;
  vector<int> f(n + 1);
  vector<int> clus(n + 1, -1);
  vector<int> g(n + 1);
  for (i = 0; i < n; i++) {
    cin >> f[i + 1];
  }
  int id = 1;
  map<int, int> h;
  for (i = 0; i < n; i++) {
    if (clus[f[i + 1]] != -1) {
      g[i + 1] = clus[f[i + 1]];
    } else {
      clus[f[i + 1]] = id;
      g[i + 1] = id;
      h[id] = f[i + 1];
      id++;
    }
  }
  bool err = false;
  for (i = 1; i < id; i++) {
    if (g[h[i]] != i) {
      err = true;
    }
  }
  if (err) {
    cout << -1;
  } else {
    cout << id - 1 << '\n';
    for (i = 0; i < n; i++) {
      cout << g[i + 1] << " ";
    }
    cout << '\n';
    for (i = 1; i < id; i++) {
      cout << h[i] << ' ';
    }
  }
  return 0;
}
