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>
const int N = 1e5 + 5;
int n, m, i, f[N], h[N], g[N], p[N];
bool have[N];
int read() {
  char c = getchar();
  int k = 0;
  for (; c < 48 || c > 57; c = getchar())
    ;
  for (; c > 47 && c < 58; c = getchar()) k = (k << 3) + (k << 1) + c - 48;
  return k;
}
void write(int x) {
  if (x > 9) write(x / 10);
  putchar(x % 10 + 48);
}
int main() {
  for (n = read(), i = 1; i <= n; i++) {
    f[i] = read();
    if (!have[f[i]]) have[f[i]] = 1, h[++m] = f[i], p[f[i]] = m;
  }
  for (i = 1; i <= n; i++) g[i] = p[f[i]];
  for (i = 1; i <= m; i++)
    if (g[h[i]] != i) {
      puts("-1");
      return 0;
    }
  write(m);
  putchar('\n');
  for (i = 1; i <= n; i++) write(g[i]), putchar(' ');
  putchar('\n');
  for (i = 1; i <= m; i++) write(h[i]), putchar(' ');
}
