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 void scf(int &x) { scanf("%d", &x); }
inline void scf(long long &x) { scanf("%I64d", &x); }
inline void scf(unsigned long long &x) { scanf("%I64u", &x); }
inline void scf(double &x) { scanf("%lf", &x); }
inline void scf(char &x) { scanf("%c", &x); }
inline void scf(char *x) { scanf("%s", x); }
const int N = 1e5 + 7;
const int mars = 0x3c3c3c3c;
typedef int Arr[N];
typedef char Err[N];
typedef bool Brr[N];
typedef long long Lrr[N];
int m, n, k, w, a, b, c, d, p, q, x, y, t, cnt, l, r, s;
Arr F, G, H;
int main() {
  int i, j;
  scf(n);
  for (i = 1; i <= n; i++) scf(F[i]);
  for (i = 1; i <= n; i++)
    if (F[i] != F[F[i]]) {
      puts("-1");
      return 0;
    }
  for (i = 1; i <= n; i++)
    if (i == F[i]) H[++m] = i, G[i] = m;
  for (i = 1; i <= n; i++)
    if (!G[i]) G[i] = G[F[i]];
  printf("%d\n", m);
  for (i = 1; i <= n; i++) printf("%d ", G[i]);
  puts("");
  for (i = 1; i <= m; i++) printf("%d ", H[i]);
}
