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>
int main(int argc, char** argv) {
  std::cin.tie(0);
  std::ios_base::sync_with_stdio(0);
  int n, m, f[111111], g[111111], h[1111111];
  std::cin >> n;
  std::set<int> set;
  for (int i = 0; i < n; i++) {
    std::cin >> f[i];
    set.insert(f[i]);
  }
  m = set.size();
  int ind = 0;
  for (const auto& i : set) h[ind++] = i;
  for (int i = 0; i < n; i++) g[i] = std::lower_bound(h, h + m, f[i]) - h + 1;
  bool valid = true;
  if (valid)
    for (int i = 0; i < m; i++)
      if (g[h[i] - 1] != i + 1) {
        valid = false;
        break;
      }
  if (valid)
    for (int i = 0; i < n; i++)
      if (h[g[i] - 1] != f[i]) {
        valid = false;
        break;
      }
  if (valid) {
    std::cout << m << std::endl;
    for (int i = 0; i < n; i++) std::cout << g[i] << ' ';
    std::cout << std::endl;
    for (int i = 0; i < m; i++) std::cout << h[i] << ' ';
    std::cout << std::endl;
  } else {
    std::cout << -1 << std::endl;
  }
  return 0;
}
