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 long long INF = 1e18;
const int mod = 1e9 + 7;
const int N = 1e5 + 10;
inline int add(int x, int y) {
  x += y;
  if (x >= mod) x -= mod;
  return x;
}
inline int mul(int x, int y) {
  x = (1LL * x * y) % mod;
  return x;
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  int n;
  cin >> n;
  vector<int> f(n + 1);
  map<int, int> M;
  for (int i = 1; i <= n; ++i) {
    cin >> f[i];
    M[f[i]];
  }
  int m = 1;
  for (auto it : M) {
    M[it.first] = m;
    m++;
  }
  vector<int> g(n + 1);
  for (int i = 1; i <= n; ++i) {
    g[i] = M[f[i]];
  }
  m--;
  vector<int> h(m + 1);
  for (int i = 1; i <= n; ++i) {
    h[g[i]] = f[i];
  }
  for (int i = 1; i <= m; ++i) {
    if (g[h[i]] != i) {
      cout << -1 << "\n";
      return 0;
    }
  }
  cout << m << "\n";
  for (int i = 1; i <= n; ++i) cout << g[i] << " ";
  cout << "\n";
  for (int i = 1; i <= m; ++i) cout << h[i] << " ";
  return 0;
}
