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 double PI = acos(-1);
const int N = 1e5 + 10;
const long long mod = 1e9 + 7;
const int inf = 0x3f3f3f3f;
const long long INF = 0x3f3f3f3f3f3f3f3fLL;
int f[N], n, m, g[N], h[N], vis[N];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  m = 0;
  memset(vis, 0, sizeof(vis));
  for (int i = 1; i <= n; i++) {
    cin >> f[i];
    if (f[i] == i) {
      h[++m] = i;
      vis[i] = m;
    }
  }
  for (int i = 1; i <= n; i++) {
    if (!vis[f[i]]) {
      cout << -1 << endl;
      return 0;
    }
  }
  cout << m << endl;
  for (int i = 1; i <= n; i++) {
    cout << vis[f[i]] << " ";
  }
  cout << endl;
  for (int i = 1; i <= m; i++) {
    cout << h[i] << " ";
  }
  cout << endl;
}