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;
using ll = long long;
const int sz = 1e5 + 10;
int f[sz], ar[sz], cnt;
int main() {
  int n, fl;
  set<int> st;
  while (cin >> n) {
    st.clear();
    cnt = 0;
    for (int i = 1; i <= n; i++) {
      cin >> f[i];
      if (st.find(f[i]) == st.end()) {
        ar[cnt++] = f[i];
      }
      st.insert(f[i]);
    }
    fl = 1;
    for (int i = 1; i <= n; i++)
      if (f[f[i]] != f[i]) fl = 0;
    if (fl) {
      sort(ar, ar + cnt);
      cout << cnt << "\n";
      for (int i = 1; i <= n; i++) {
        cout << (lower_bound(ar, ar + cnt, f[i]) - ar) + 1 << " ";
      }
      cout << "\n";
      for (int x : st) cout << x << " ";
      cout << "\n";
    } else {
      cout << "-1\n";
    }
  }
  return 0;
}