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;
int h[100005], f[100005], g[100005];
int nilai[1000005];
int ukur = 1;
int main() {
  int n;
  cin >> n;
  memset(g, -1, sizeof(g));
  for (int tmp = 1; tmp <= n; tmp++) {
    cin >> f[tmp];
    if (f[tmp] == tmp) {
      h[ukur] = tmp;
      g[tmp] = ukur;
      ukur++;
    }
  }
  for (int tmp = 1; tmp <= n; tmp++) {
    if (f[f[tmp]] != f[tmp]) {
      cout << "-1\n";
      return 0;
    }
  }
  for (int tmp = 1; tmp <= n; tmp++) {
    if (g[tmp] == -1) {
      g[tmp] = g[f[tmp]];
    }
  }
  cout << ukur - 1 << "\n";
  for (int tmp = 1; tmp <= n; tmp++) {
    if (tmp != 1) cout << " ";
    cout << g[tmp];
  }
  cout << "\n";
  for (int tmp = 1; tmp < ukur; tmp++) {
    if (tmp != 1) cout << " ";
    cout << h[tmp];
  }
  cout << "\n";
}
