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;
using LD = long double;
int main() {
  cin.tie(nullptr);
  ios_base::sync_with_stdio(false);
  cout.setf(ios_base::fixed);
  cout.precision(28);
  int n;
  cin >> n;
  vector<int> x(n);
  for (int i = 0; i < n; ++i) {
    cin >> x[i];
    x[i]--;
  }
  vector<int> h, g(n, -1);
  vector<int> zn(n, -1);
  for (int i = 0; i < n; ++i) {
    if (zn[x[i]] == -1) {
      h.push_back(x[i]);
      zn[x[i]] = h.size() - 1;
    }
    g[i] = zn[x[i]];
  }
  bool ok = true;
  for (int i = 0; i < h.size(); ++i) {
    if (g[h[i]] != i) {
      ok = false;
    }
  }
  if (!ok) {
    cout << -1;
    return 0;
  }
  cout << h.size() << "\n";
  for (int i = 0; i < n; ++i) {
    cout << g[i] + 1 << " ";
  }
  cout << "\n";
  for (int i = 0; i < h.size(); ++i) {
    cout << h[i] + 1 << " ";
  }
  return 0;
}
