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 N = 100005;
int n, a, b, c, f[N];
vector<int> s;
int inv[N], g[N], h[N];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", f + i);
    if (f[i] == i) s.push_back(i);
  }
  bool ok = true;
  for (int i = 1; i <= n; ++i)
    if (f[f[i]] != f[i]) ok = false;
  int len = s.size();
  if (len == 0 or not ok) {
    puts("-1");
    return 0;
  }
  printf("%d\n", len);
  int t = 1;
  for (int it : s) inv[it] = t++;
  for (int i = 1; i <= n; ++i) g[i] = inv[f[i]];
  for (int i = 1; i <= n; ++i) printf("%d ", g[i]);
  puts("");
  for (int i = 1; i <= len; ++i) printf("%d ", s[i - 1]);
  puts("");
  return 0;
}