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 maxn = 200000 + 23;
int n, F[maxn], num[maxn], vis[maxn], a[maxn], size;
bool mark;
int find(int k) { return (F[k] == k) ? (k) : (F[k] = find(F[k])); }
void unit(int x, int y) {
  int fx = find(x), fy = find(y);
  if (fx == fy)
    return;
  else
    F[fx] = fy;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]), F[i] = i;
  for (int i = 1; i <= n; i++) unit(i, a[i]);
  size = 0;
  memset(vis, 0, sizeof(vis));
  for (int i = 1; i <= n; i++) {
    if (!vis[find(i)]) vis[find(i)] = ++size;
    num[i] = vis[find(i)];
  }
  mark = 1;
  memset(vis, 0, sizeof(vis));
  for (int i = 1; i <= n; i++) {
    if (vis[num[i]] == 0)
      vis[num[i]] = a[i];
    else {
      if (vis[num[i]] != a[i]) {
        mark = 0;
        break;
      }
    }
  }
  if (!mark)
    printf("%d\n", -1);
  else {
    printf("%d\n", size);
    for (int i = 1; i <= n; i++)
      printf("%d%c", num[i], (i == n) ? ('\n') : (' '));
    for (int i = 1; i <= size; i++)
      printf("%d%c", vis[i], (i == size) ? ('\n') : (' '));
  }
  return 0;
}
