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;
long long gcd(long long n1, long long n2) {
  if (n1 % n2 == 0) return n2;
  return gcd(n2, n1 % n2);
}
long long powmod(long long base, long long exponent) {
  long long ans = 1;
  while (exponent) {
    if (exponent & 1) ans = (ans * base) % 10000000000000007LL;
    base = (base * base) % 10000000000000007LL;
    exponent /= 2;
  }
  return ans;
}
int cnt[200];
bool flag = true;
int g[300005];
int f[300005];
int h[300005];
vector<int> v[300005];
void DFS(int node, int k, int tt) {
  if (g[node] > 0)
    return;
  else {
    g[node] = k;
    if (f[node] != tt) {
      flag = false;
    }
  }
  for (int i = 0; i < v[node].size(); i++) DFS(v[node][i], k, tt);
}
int main() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> f[i];
  for (int i = 1; i <= n; i++) {
    v[i].push_back(f[i]);
    v[f[i]].push_back(i);
  }
  int k = 1;
  for (int i = 1; i <= n; i++)
    if (g[i] == 0) {
      h[k] = f[i];
      DFS(i, k, f[i]), k++;
    }
  if (flag == false) {
    cout << -1 << '\n';
    return 0;
  }
  k--;
  cout << k << '\n';
  for (int i = 1; i <= n; i++) cout << g[i] << " ";
  cout << '\n';
  for (int i = 1; i <= k; i++) cout << h[i] << " ";
  cout << '\n';
}
