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;
int n, m;
int a[100005], h[100005], g[100005];
bool flag = 1;
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  for (; ch < '0' || ch > '9'; ch = getchar())
    if (ch == '-') f = -1;
  for (; ch >= '0' && ch <= '9'; ch = getchar()) x = x * 10 + ch - '0';
  return x * f;
}
inline void print(int x) {
  if (x < 0) {
    putchar('-');
    x = -x;
  }
  if (x >= 10) print(x / 10);
  putchar(x % 10 + '0');
}
int main() {
  n = read();
  for (int i = 1; i <= n; i++) {
    a[i] = read();
    if (a[i] == i) g[i] = ++m, h[m] = i, flag = 1;
  }
  if (!flag) {
    puts("-1");
    return 0;
  }
  for (int i = 1; i <= n; i++)
    if (a[i] != i) {
      if (a[a[i]] != a[i]) {
        puts("-1");
        return 0;
      }
      g[i] = g[a[i]];
    }
  printf("%d\n", m);
  for (int i = 1; i <= n; i++) print(g[i]), putchar(' ');
  puts("");
  for (int i = 1; i <= m; i++) print(h[i]), putchar(' ');
  puts("");
  return 0;
}
