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 inf = 0x7f7f7f7f;
template <typename t>
t read() {
  t x = 0;
  int f = 1;
  char c = getchar();
  while (c > '9' || c < '0') f = c == '-' ? -1 : 1, c = getchar();
  while (c >= '0' && c <= '9') x = x * 10 + c - 48, c = getchar();
  return x * f;
}
template <typename t>
void write(t x) {
  if (x < 0) {
    putchar('-'), write(-x);
    return;
  }
  if (x >= 10) write(x / 10);
  putchar(x % 10 + 48);
}
const int maxn = 1e6 + 10;
int f[maxn], g[maxn], h[maxn], p[maxn], n, m;
int main() {
  n = read<int>();
  for (int i = 1; i <= n; i++) {
    f[i] = read<int>();
    if (i == f[i]) h[p[i] = ++m] = i;
  }
  for (int i = 1; i <= n; i++) {
    if (!p[f[i]]) return puts("-1"), 0;
    g[i] = p[f[i]];
  }
  printf("%d\n", m);
  for (int i = 1; i <= n; i++) printf("%d ", g[i]);
  puts("");
  for (int i = 1; i <= m; i++) printf("%d ", h[i]);
  return 0;
}
