Problem: 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#pragma comment(linker, "/stack:16777216")
using namespace std;
const int INF = 1000000000;
const int BASE = 1000000007;
const int MAX = 100007;
const int MAX2 = 7777;
const int MAXE = 100000;
const int ADD = 1000000;
const int MOD = 1000000007;
const int CNT = 800;
int main() {
  int n;
  cin >> n;
  vector<int> f(n);
  set<int> s;
  for (int i = (0); i < (n); i++) {
    scanf("%d", &f[i]);
    --f[i];
    if (f[i] == i) {
      s.insert(i);
    }
  }
  for (int i = (0); i < (n); i++) {
    if (s.find(f[i]) == s.end()) {
      cout << -1 << endl;
      return 0;
    }
  }
  int m = (int)s.size();
  map<int, int> M;
  vector<int> h;
  for (set<int>::iterator it = s.begin(); it != s.end(); ++it) {
    M[*it] = (int)h.size();
    h.push_back(*it);
  }
  vector<int> g;
  for (int i = (0); i < (n); i++) {
    g.push_back(M[f[i]]);
  }
  cout << m << endl;
  for (int i = (0); i < (n); i++) {
    printf("%d ", g[i] + 1);
  }
  cout << endl;
  for (int i = (0); i < (m); i++) {
    printf("%d ", h[i] + 1);
  }
  cout << endl;
  return 0;
}