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;
constexpr int MAXN = 1e6;
int n, f[MAXN + 5], g[MAXN + 5], h[MAXN + 5], m;
vector<int> pimg[MAXN + 5];
bool isValid;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    cin >> f[i];
    pimg[f[i]].push_back(i);
  }
  isValid = true;
  for (int i = 1; i <= n; ++i) {
    if (!pimg[i].empty()) {
      if (f[i] != i) isValid = false;
      ++m;
      h[m] = i;
      for (auto& p : pimg[i]) {
        g[p] = m;
      }
    }
  }
  if (isValid) {
    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] << ' ';
    cout << '\n';
  } else {
    cout << -1 << '\n';
  }
}