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 namespace std::chrono;
const int N = 300009;
const int LOG = 20;
int as(int node, int k, vector<vector<int>> &anc) {
  if (k == 0) return node;
  for (int i = 0; i < LOG; i++) {
    if (k & (1 << i)) {
      node = anc[node][i];
    }
  }
  return node;
}
void solve() {
  int n;
  cin >> n;
  vector<int> a(n);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    a[i]--;
  }
  vector<vector<int>> x(n);
  for (int i = 0; i < n; i++) {
    x[a[i]].push_back(i);
  }
  for (int i = 0; i < n; i++) {
    sort(x[i].begin(), x[i].end());
  }
  bool valid = true;
  int count = 0;
  vector<int> g(n);
  vector<int> h;
  for (int i = 0; i < n; i++) {
    if (x[i].size() != 0 && !binary_search(x[i].begin(), x[i].end(), i)) {
      valid = false;
      break;
    } else if (x[i].size() != 0) {
      h.push_back(i);
      for (auto c : x[i]) g[c] = count;
      count++;
    }
  }
  if (!valid) {
    cout << -1 << "\n";
    return;
  }
  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 << " ";
  }
  cout << "\n";
}
int main() {
  auto start = high_resolution_clock::now();
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  for (int i = 0; i < t; i++) {
    solve();
  }
  auto stop = high_resolution_clock::now();
  auto duration = duration_cast<milliseconds>(stop - start);
}
