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;
template <typename T>
void _R(T &x) {
  cin >> x;
}
void _R(int &x) { scanf("%d", &x); }
void _R(long long &x) { scanf("%" PRId64, &x); }
void _R(double &x) { scanf("%lf", &x); }
void _R(char &x) { scanf(" %c", &x); }
void _R(char *x) { scanf("%s", x); }
void R() {}
template <typename T, typename... U>
void R(T &head, U &...tail) {
  _R(head);
  R(tail...);
}
template <typename T>
void _W(const T &x) {
  cout << x;
}
void _W(const int &x) { printf("%d", x); }
template <typename T>
void _W(const vector<T> &x) {
  for (auto i = x.cbegin(); i != x.cend(); i++) {
    if (i != x.cbegin()) putchar(' ');
    _W(*i);
  }
}
void W() {}
template <typename T, typename... U>
void W(const T &head, const U &...tail) {
  _W(head);
  putchar(sizeof...(tail) ? ' ' : '\n');
  W(tail...);
}
const int N = 1e5 + 10;
int n, f[N], g[N], h[N], z[N];
int main() {
  R(n);
  for (int i = (1); i <= int(n); i++) R(f[i]);
  int m = 0;
  for (int i = (1); i <= int(n); i++)
    if (f[i] == i) {
      m++;
      g[i] = m;
      h[m] = i;
      z[i] = m;
    }
  if (m == 0) {
    W(-1);
    return 0;
  }
  for (int i = (1); i <= int(n); i++)
    if (f[i] != i) {
      if (z[f[i]] == 0) {
        W(-1);
        return 0;
      }
      g[i] = z[f[i]];
    }
  for (int i = (1); i <= int(m); i++) assert(g[h[i]] = i);
  for (int i = (1); i <= int(n); i++) assert(h[g[i]] == f[i]);
  W(m);
  W(vector<int>(g + 1, g + n + 1));
  W(vector<int>(h + 1, h + m + 1));
  return 0;
}
