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;
const long long N = 1e6;
long long f[N], g[N], h[N];
int32_t main() {
  long long n = ({
    long long x;
    scanf("%lld", &x);
    x;
  });
  set<long long> st;
  for (long long i = 1; i <= n; i++) {
    f[i] = ({
      long long x;
      scanf("%lld", &x);
      x;
    });
    st.insert(f[i]);
  }
  vector<long long> cc(st.begin(), st.end());
  long long m = cc.size();
  for (long long i = 1; i <= m; i++) h[i] = cc[i - 1];
  for (long long i = 1; i <= n; i++)
    g[i] = lower_bound((cc).begin(), (cc).end(), f[i]) - cc.begin() + 1;
  {
    for (long long i = 1; i <= m; i++) {
      if (g[h[i]] != i) {
        cout << -1 << endl;
        return 0;
      }
    }
    for (long long i = 1; i <= n; i++) {
      if (h[g[i]] != f[i]) {
        cout << -1 << endl;
        return 0;
      }
    }
  }
  cout << m << endl;
  for (long long i = 1; i <= n; i++) cout << g[i] << " ";
  cout << endl;
  for (long long i = 1; i <= m; i++) cout << h[i] << " ";
  cout << endl;
}
