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;
long long dp[int(3e5 + 11)];
long long fx[int(3e5 + 11)];
long long fe[int(3e5 + 11)];
long long pas[int(3e5 + 11)];
long long pp[int(3e5 + 11)];
long long f[int(3e5 + 11)];
vector<long long> v[int(3e5 + 11)];
vector<long long> cl[int(3e5 + 11)];
long long a[int(3e5 + 11)];
int main() {
  long long n, b, m = 0;
  cin >> n;
  long long k = 0;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    if (i == a[i]) {
      pp[i] = k;
      pas[k] = i;
      fe[a[i]] = 1;
      k++;
    }
    if (!fx[a[i]]) m++;
    fx[a[i]] = 1;
  }
  for (int i = 0; i <= n; i++) {
    if (fx[i] && !fe[i]) {
      cout << -1;
      return 0;
    }
  }
  cout << m << endl;
  for (int i = 1; i <= n; i++) {
    cout << pp[a[i]] + 1 << " ";
  }
  cout << endl;
  for (int i = 0; i < m; i++) {
    cout << pas[i] << " ";
  }
  return 0;
}