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 arr[100009];
int main() {
  bool flag = true;
  int n, f[100009], m, g[100009], h[100009];
  unordered_map<int, vector<int>> ma;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> f[i];
    if (ma.find(f[i]) != ma.end()) {
      ma[f[i]].push_back(i);
    } else {
      vector<int> v;
      v.push_back(i);
      ma.emplace(f[i], v);
    }
  }
  m = ma.size();
  for (unordered_map<int, vector<int>>::iterator i = ma.begin(); i != ma.end();
       i++) {
    bool b = false;
    for (int j = 0; j < i->second.size(); j++) {
      if (i->second[j] == i->first) {
        b = true;
        break;
      }
    }
    if (b == false) {
      flag = false;
      cout << -1;
      return 0;
    }
  }
  int num = 1;
  for (unordered_map<int, vector<int>>::iterator i = ma.begin(); i != ma.end();
       i++) {
    for (int j = 0; j < i->second.size(); j++) {
      g[i->second[j]] = num;
    }
    h[num] = i->first;
    num++;
  }
  cout << m << endl;
  for (int i = 1; i <= n; i++) {
    cout << g[i] << ' ';
  }
  cout << endl;
  for (int j = 1; j <= m; j++) {
    cout << h[j] << ' ';
  }
  return 0;
}
