Problem: 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int n, f[100009];
bool m1[100009], m2[100009];
int m, id[100009], rid[100009];
bool proc() {
  for (int i = 0; i < n; i++)
    if (m1[i] && !m2[i]) return 0;
  return 1;
}
int main() {
  while (scanf("%d", &n) == 1) {
    m = 0;
    memset(m1, 0, sizeof(m1));
    memset(m2, 0, sizeof(m2));
    memset(id, -1, sizeof(id));
    for (int i = 0; i < n; i++) {
      scanf("%d", &f[i]);
      f[i]--;
      if (!m1[f[i]]) {
        m1[f[i]] = 1;
        rid[m] = f[i];
        id[f[i]] = m++;
      }
      if (i == f[i]) m2[f[i]] = 1;
    }
    if (!proc())
      puts("-1");
    else {
      printf("%d\n", m);
      bool fp = 1;
      for (int i = 0; i < n; i++) {
        if (fp)
          fp = 0;
        else
          printf(" ");
        printf("%d", id[f[i]] + 1);
      }
      puts("");
      fp = 1;
      for (int i = 0; i < m; i++) {
        if (fp)
          fp = 0;
        else
          printf(" ");
        printf("%d", rid[i] + 1);
      }
      puts("");
    }
  }
  return 0;
}