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 main() {
  int n;
  cin >> n;
  int f[n];
  for (int i = 0; i < n; i++) cin >> f[i];
  map<int, int> ma;
  map<int, vector<int> > G;
  int m;
  for (int i = 0; i < n; i++) {
    f[i]--;
    m = ma.size();
    if (ma.count(f[i])) m = ma[f[i]];
    ma[f[i]] = m;
    G[m].push_back(i);
  }
  m = ma.size();
  int g[n], h[m];
  memset(h, -1, sizeof(h));
  for (int i = 0; i < n; i++) g[i] = ma[f[i]];
  for (int i = 0; i < n; i++) h[g[i]] = f[i];
  bool a = 1;
  for (int i = 0; i < m; i++) a &= g[h[i]] == i;
  for (int i = 0; i < n; i++) a &= h[g[i]] == f[i];
  if (a) {
    cout << m << endl;
    for (int i = 0; i < n; i++) cout << g[i] + 1 << " \n"[i == n - 1];
    for (int i = 0; i < m; i++) cout << h[i] + 1 << " \n"[i == m - 1];
  } else
    cout << -1 << endl;
  return 0;
}
