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() {
  size_t n;
  std::cin >> n;
  std::vector<size_t> f(n + 1);
  for (size_t x = 1; x <= n; ++x) {
    std::cin >> f[x];
  }
  size_t m = 0;
  std::vector<std::vector<size_t>> f_preimage(n + 1);
  for (size_t x = 1; x <= n; ++x) {
    auto& xs = f_preimage[f[x]];
    if (xs.empty()) {
      ++m;
    }
    xs.push_back(x);
  }
  std::vector<size_t> h(m + 1);
  std::vector<size_t> g(n + 1);
  size_t hx = 0;
  for (size_t fy = 1; fy <= n; ++fy) {
    if (!f_preimage[fy].empty()) {
      h[++hx] = fy;
      bool valid = false;
      for (size_t fx : f_preimage[fy]) {
        g[fx] = hx;
        if (fx == fy) {
          valid = true;
        }
      }
      if (!valid) {
        std::cout << -1 << std::endl;
        return 0;
      }
    }
  }
  std::cout << m << std::endl;
  for (size_t x = 1; x <= n; ++x) {
    std::cout << g[x] << " ";
  }
  std::cout << std::endl;
  for (size_t x = 1; x <= m; ++x) {
    std::cout << h[x] << " ";
  }
  std::cout << std::endl;
  return 0;
}
