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;
inline int toInt(string s) {
  int v;
  istringstream sin(s);
  sin >> v;
  return v;
}
template <class T>
inline string toString(T x) {
  ostringstream sout;
  sout << x;
  return sout.str();
}
int a[111111];
int c[111111];
int g[111111];
int h[111111];
int main() {
  int n, k = 0;
  scanf("%d", &n);
  for (int i = (0); i < (n); ++i) {
    scanf("%d", &a[i]);
    a[i]--;
  }
  for (int i = (0); i < (n); ++i)
    if (a[a[i]] != a[i]) k = -1;
  if (k == -1) {
    printf("-1\n");
    return 0;
  }
  memset(g, -1, sizeof(g));
  for (int i = (0); i < (n); ++i) {
    if (g[a[i]] == -1) {
      g[i] = g[a[i]] = k;
      h[k] = a[i];
      k++;
    }
    g[i] = g[a[i]];
  }
  printf("%d\n", k);
  for (int i = (0); i < (n); ++i) printf("%d ", g[i] + 1);
  printf("\n");
  for (int i = (0); i < (k); ++i) printf("%d ", h[i] + 1);
  printf("\n");
  return 0;
}
