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 int N = 1000006;
int f[N], g[N], h[N];
int p[N], val[N];
int root(int i) { return p[i] = p[i] == i ? i : root(p[i]); }
void merge(int i, int j) {
  i = root(i);
  j = root(j);
  if (i > j) {
    swap(i, j);
  }
  p[j] = i;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", f + i);
  }
  for (int i = 0; i < N; ++i) {
    p[i] = i;
    val[i] = i;
  }
  for (int x = 1; x <= n; ++x) {
    merge(f[x], x);
  }
  for (int x = 1; x <= n; ++x) {
    g[x] = root(x);
  }
  for (int x = 1; x <= n; ++x) {
    if (h[g[x]] && h[g[x]] != f[x]) {
      puts("-1");
      return 0;
    }
    h[g[x]] = f[x];
  }
  int m = n;
  queue<int> empty;
  for (int x = 1; x <= n; ++x) {
    if (!empty.empty() && h[x]) {
      int new_x = empty.front();
      empty.pop();
      val[g[h[x]]] = new_x;
      h[new_x] = h[x];
      empty.push(x);
    }
    if (h[x] == 0) {
      empty.push(x);
      --m;
    }
  }
  for (int x = 1; x <= n; ++x) {
    g[x] = val[g[x]];
  }
  printf("%d\n", m);
  for (int i = 1; i <= n; ++i) {
    printf("%d%c", g[i], " \n"[i == n]);
  }
  for (int i = 1; i <= m; ++i) {
    printf("%d%c", h[i], " \n"[i == n]);
  }
}
