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;
vector<int> v[1000005];
int vis[1000005];
int f[1000005];
int g[1000001];
int h[1000001];
int main() {
  int n;
  cin >> n;
  vector<int> v2;
  for (int i = 1; i <= n; i++) {
    cin >> f[i];
    v[f[i]].push_back(i);
    if (!vis[f[i]]) {
      v2.push_back(f[i]);
      vis[f[i]] = 1;
    }
  }
  int cnt = 0;
  for (int i = 1; i <= 100000; i++) {
    if (v[i].size() > 0) {
      cnt++;
    }
  }
  int mms = 1;
  for (int i = 0; i < v2.size(); i++) {
    int p = v2[i];
    for (int j = 0; j < v[p].size(); j++) {
      g[v[p][j]] = mms;
    }
    h[mms] = f[v[p][0]];
    mms++;
  }
  int fl = 0;
  for (int i = 1; i <= n; i++) {
    if (h[g[i]] == f[i]) {
      continue;
    } else
      fl = 1;
  }
  for (int i = 1; i <= cnt; i++) {
    if (g[h[i]] == i) {
      continue;
    } else
      fl = 1;
  }
  if (fl == 1) {
    cout << -1 << endl;
    return 0;
  }
  cout << cnt << endl;
  for (int i = 1; i <= n; i++) {
    cout << g[i] << " ";
  }
  cout << endl;
  for (int i = 1; i < mms; i++) {
    cout << h[i] << " ";
  }
  cout << endl;
}