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;
long long MOD = 1e9 + 7;
int b[100001];
int main() {
  ios_base::sync_with_stdio(0);
  int n;
  cin >> n;
  int a[n + 1];
  for (int i = 0; i < n; ++i) {
    cin >> a[i + 1];
    b[a[i + 1]] = 1;
  }
  vector<int> v;
  for (int i = 1; i <= n; ++i) {
    if (b[i]) {
      v.push_back(i);
      if (a[i] != i) {
        cout << -1 << endl;
        return 0;
        ;
      }
    }
  }
  int m = v.size();
  cout << m << endl;
  int g[n + 1], h[m + 1];
  unordered_map<int, int> mm;
  for (int i = 0; i < m; ++i) {
    h[i + 1] = v[i];
    mm[v[i]] = i + 1;
  }
  for (int i = 0; i < n; ++i) {
    int x = a[i + 1];
    x = mm[x];
    g[i + 1] = x;
  }
  for (int i = 0; i < n; ++i) {
    cout << g[i + 1] << " ";
  }
  cout << endl;
  for (int i = 0; i < m; ++i) {
    cout << h[i + 1] << " ";
  }
  cout << endl;
  return 0;
}
