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 <class T, class U>
void chmin(T &t, U f) {
  if (t > f) t = f;
}
template <class T, class U>
void chmax(T &t, U f) {
  if (t < f) t = f;
}
long long N;
long long to[111111];
bool ok[111111];
vector<long long> g, h;
signed main() {
  scanf("%lld", &N);
  for (long long i = 0; i < (N); i++) scanf("%lld", &to[i]), to[i]--;
  for (long long i = 0; i < (N); i++)
    if (to[i] == i) ok[i] = true;
  for (long long i = 0; i < (N); i++)
    if (!ok[i] && !ok[to[i]]) {
      puts("-1");
      return 0;
    }
  g.resize(N);
  for (long long i = 0; i < (N); i++) {
    if (ok[i]) {
      g[i] = h.size();
      h.push_back(i);
    }
  }
  for (long long i = 0; i < (N); i++) {
    if (!ok[i]) {
      g[i] = g[to[i]];
    }
  }
  printf("%lld\n", (long long)h.size());
  for (long long i = 0; i < (N); i++) printf("%lld ", g[i] + 1);
  puts("");
  for (long long i = 0; i < (h.size()); i++) printf("%lld ", h[i] + 1);
  puts("");
  return 0;
}
