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;
const int INF = int(1e9) + 1;
int main() {
  ios_base::sync_with_stdio(false);
  int n;
  cin >> n;
  vector<int> f(n);
  vector<vector<int> > S(n);
  for (int i = 0; i < n; ++i) {
    int x;
    cin >> x;
    --x;
    f[i] = x;
    S[x].push_back(i);
    S[x].push_back(x);
  }
  vector<int> g(n, -1);
  bool used = false;
  int c = 0;
  for (int i = 0; i < n; ++i) {
    used = false;
    for (int j = 0; j < S[i].size(); ++j) {
      if (g[S[i][j]] != -1 and g[S[i][j]] != c) {
        cout << -1 << endl;
        return 0;
      }
      g[S[i][j]] = c;
      if (!used) used = true;
    }
    if (used) ++c;
  }
  vector<int> h(c, -1);
  for (int i = 0; i < n; ++i) {
    if (g[i] != -1) h[g[i]] = f[i];
  }
  cout << c << endl;
  for (int i = 0; i < n; ++i) {
    cout << (i == 0 ? "" : " ") << g[i] + 1;
  }
  cout << endl;
  for (int i = 0; i < c; ++i) {
    cout << (i == 0 ? "" : " ") << (h[i] == -1 ? 1 : h[i] + 1);
  }
  cout << endl;
}
