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;
inline int read() {
  int s = 0, t = 1;
  char c = getchar();
  while (!isdigit(c)) {
    if (c == '-') t = -1;
    c = getchar();
  }
  while (isdigit(c)) s = s * 10 + c - 48, c = getchar();
  return s * t;
}
const int N = 1e5 + 5;
char c[N];
bool vis[N];
int q[N], v[N], g[N];
int main() {
  int n = read(), tot = 0, r = 0;
  for (register int i = 1; i <= n; i++) {
    v[i] = read();
    if (!vis[v[i]]) tot++, vis[v[i]] = 1;
    if (v[i] == i) q[++r] = i;
  }
  if (tot != r) return puts("-1"), 0;
  sort(q + 1, q + r + 1);
  for (register int i = 1; i <= n; i++) {
    g[i] = lower_bound(q + 1, q + r + 1, v[i]) - q;
  }
  cout << r << endl;
  for (register int i = 1; i <= n; i++) cout << g[i] << " ";
  cout << endl;
  for (register int i = 1; i <= r; i++) cout << q[i] << " ";
  cout << endl;
  return 0;
}
