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() {
  long n;
  scanf("%ld", &n);
  std::vector<long> f(n + 1, 0), g(n + 1, 0), h(n + 1, 0), vis(n + 1, 0);
  long m(0);
  for (int p = 1; p <= n; p++) {
    scanf("%ld", &f[p]);
    if (f[p] == p) {
      h[++m] = p;
      g[p] = m;
      vis[p] = true;
    }
  }
  bool possible(true);
  for (int p = 1; p <= n; p++) {
    if (!vis[p]) {
      if (!vis[f[p]]) {
        possible = 0;
        break;
      } else
        g[p] = g[f[p]];
    }
  }
  if (!possible) {
    puts("-1");
  } else {
    printf("%ld\n", m);
    for (int p = 1; p <= n; p++) {
      printf("%ld ", g[p]);
    };
    puts("");
    for (int p = 1; p <= m; p++) {
      printf("%ld ", h[p]);
    };
    puts("");
  }
  return 0;
}
