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
#include <bits/stdc++.h>
using namespace std;
int n;
int f[100001];
int g[100001];
int h[100001];
int g_inv[100001];
int h_inv[100001];
bool seen[100001];
int ok = true;
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) {
    int tmp;
    cin >> tmp;
    f[i] = tmp - 1;
    seen[f[i]] = true;
  }
  int m = 0;
  for (int i = 0; i < n; i++) {
    if (seen[i]) {
      h[m] = i;
      h_inv[i] = m;
      m++;
    }
  }
  for (int i = 0; i < n; i++) {
    g[i] = -1;
  }
  for (int i = 0; i < m; i++) {
    int hx = h[i];
    if (g[hx] != -1 && g[hx] != i) {
      ok = false;
    }
    g[hx] = i;
  }
  for (int i = 0; i < n; i++) {
    int gx = h_inv[f[i]];
    if (g[i] != -1 && g[i] != gx) {
      ok = false;
    }
    g[i] = gx;
  }
  for (int i = 0; i < n; i++) {
    if (g[i] == -1) g[i] = 0;
  }
  if (!ok) {
    cout << "-1";
    return 0;
  }
  cout << m << "\n";
  for (int i = 0; i < n - 1; i++) {
    cout << g[i] + 1 << " ";
  }
  cout << g[n - 1] + 1 << "\n";
  for (int i = 0; i < m - 1; i++) {
    cout << h[i] + 1 << " ";
  }
  cout << h[m - 1] + 1 << "\n";
}
