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;
const int maxll = 1e05 + 10;
int n, m(0);
bool aprd[maxll];
int f[maxll];
int h[maxll];
int g[maxll];
int cl[maxll];
map<int, int> RFC;
int find_f(int x) {
  if (cl[x] != x) cl[x] = find_f(cl[x]);
  return cl[x];
}
int main() {
  memset(aprd, false, sizeof(aprd));
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &f[i]);
    cl[i] = i;
  }
  int cnt = 0;
  for (int i = 1; i <= n; i++) {
    if (f[i] != f[f[i]]) {
      cout << "-1";
      return 0;
    }
    int x = find_f(f[i]);
    int y = find_f(i);
    int tmp = max(x, y);
    int tmp2 = min(x, y);
    cl[tmp] = tmp2;
  }
  for (int i = 1; i <= n; i++) {
    find_f(i);
    if (!aprd[cl[i]]) {
      aprd[cl[i]] = true;
      g[i] = ++cnt;
      h[cnt] = f[i];
    } else {
      g[i] = g[cl[i]];
    }
  }
  cout << cnt << endl;
  for (int i = 1; i <= n; i++) printf("%d ", g[i]);
  printf("\n");
  for (int i = 1; i <= cnt; i++) printf("%d ", h[i]);
  return 0;
}
