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;
const int INF = 1e9 + 7;
vector<int> v[100005];
int g[100005], h[100005], f[100005], id[100005];
set<int> H;
int main() {
  int n, m = 0;
  scanf("%d", &n);
  for (int k = 1; k <= (n); k++) scanf("%d", &f[k]);
  for (int k = 1; k <= (n); k++) {
    if (f[f[k]] != f[k]) {
      printf("-1");
      return 0;
    }
    H.insert(f[k]);
  }
  for (set<int>::iterator it = H.begin(); it != H.end(); it++) {
    h[++m] = *it;
    id[*it] = m;
  }
  for (int k = 1; k <= (n); k++) g[k] = id[f[k]];
  printf("%d\n", m);
  for (int k = 1; k <= (n); k++) printf("%d ", g[k]);
  putchar('\n');
  for (int k = 1; k <= (m); k++) printf("%d ", h[k]);
  return 0;
}