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;
class UnionFind {
 private:
  vector<int> p, rank;
  int numSets;

 public:
  UnionFind(int N) {
    rank.assign(N, 0);
    p.assign(N, 0);
    for (int i = 0; i < N; ++i) {
      p[i] = i;
    }
    numSets = N;
  }
  int findSet(int i) { return (p[i] == i) ? i : (p[i] = findSet(p[i])); }
  bool isSameSet(int i, int j) { return findSet(i) == findSet(j); }
  void unionSet(int i, int j) {
    if (!isSameSet(i, j)) {
      --numSets;
      int x = findSet(i), y = findSet(j);
      p[x] = y;
    }
  }
  int numDisjointSets() { return numSets; }
};
class TaskD {
 public:
  void solve(istream& in, ostream& out) {
    int n;
    in >> n;
    int m = 0;
    vector<int> f(n + 1), g(n + 1, -1), h(1), invh(n + 1, -1);
    for (int i = 1; i <= n; ++i) {
      in >> f[i];
      if (i == f[i]) {
        h.push_back(f[i]);
        invh[f[i]] = h.size() - 1;
      }
    }
    for (int i = 1; i <= n; ++i) {
      if (invh[f[i]] != -1) {
        g[i] = invh[f[i]];
      } else {
        out << -1 << "\n";
        return;
      }
    }
    m = h.size() - 1;
    out << m << "\n";
    for (int i = 1; i <= n; ++i) {
      out << g[i] << " ";
    }
    out << "\n";
    for (int i = 1; i <= m; ++i) {
      out << h[i] << " ";
    }
    out << "\n";
  }
};
int main() {
  std::ios::sync_with_stdio(false);
  TaskD solver;
  std::istream& in(std::cin);
  std::ostream& out(std::cout);
  solver.solve(in, out);
  return 0;
}
