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 int N = 1e5 + 10;
struct Node {
  int x, fx;
  bool operator<(const Node& ano) const { return fx < ano.x; }
} a[N];
int g[N], h[N];
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    a[i].x = i + 1;
    cin >> a[i].fx;
  }
  memset(g, -1, sizeof(g));
  memset(h, -1, sizeof(h));
  int cur = 1;
  for (int it = 0; it < n; it++) {
    if (g[a[it].fx] != -1) {
      if (h[g[a[it].fx]] != a[it].fx) {
        cout << "-1\n";
        return 0;
      }
      g[a[it].x] = g[a[it].fx];
    } else if (g[a[it].x] != -1) {
      if (h[g[a[it].x]] != a[it].fx) {
        cout << "-1\n";
        return 0;
      }
      g[a[it].x] = g[a[it].fx];
    } else {
      g[a[it].x] = cur++;
      h[g[a[it].x]] = a[it].fx;
      g[h[g[a[it].x]]] = g[a[it].x];
    }
  }
  cout << cur - 1 << endl;
  for (int i = 1; i <= n; i++) cout << g[i] << (i == n ? "\n" : " ");
  for (int i = 1; i < cur; i++) {
    cout << h[i] << (i == cur - 1 ? "\n" : " ");
  }
}