Problem: 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int n, m, k, tmp, id;
long long ans;
int f[(int)(1e6 + 7)], g[(int)(1e6 + 7)], h[(int)(1e6 + 7)], xh[(int)(1e6 + 7)];
void them(int vt) {
  int gt = f[vt];
  if (xh[gt] == -1) {
    xh[gt] = ++id;
    h[id] = gt;
  }
  g[vt] = xh[gt];
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &f[i]);
    xh[f[i]] = -1;
  }
  for (int i = 1; i <= n; i++) {
    if (xh[i] != 0 && i != f[i]) {
      cout << -1;
      return 0;
    }
    them(i);
  }
  printf("%d\n", id);
  for (int i = 1; i <= n; i++) printf("%d ", g[i]);
  printf("\n");
  for (int i = 1; i <= id; i++) printf("%d ", h[i]);
  return 0;
}