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;
template <typename T>
inline bool chkmin(T &a, const T &b) {
  return a > b ? a = b, 1 : 0;
}
template <typename T>
inline bool chkmax(T &a, const T &b) {
  return a < b ? a = b, 1 : 0;
}
const int oo = 0x3f3f3f3f;
const int maxn = 100100;
int n, m;
int a[maxn + 5];
int b[maxn + 5];
int c[maxn + 5];
int id[maxn + 5];
int main() {
  scanf("%d", &n);
  memset(id, -1, sizeof id);
  memset(c, -1, sizeof c);
  m = 0;
  for (int i = (0), i_end_ = (n); i < i_end_; ++i) {
    scanf("%d", a + i), --a[i];
    if (!~id[a[i]]) id[a[i]] = m++;
    b[i] = id[a[i]];
    if (a[i] == i) c[id[a[i]]] = i;
  }
  for (int i = (0), i_end_ = (m); i < i_end_; ++i)
    if (!~c[i]) {
      printf("-1\n");
      return 0;
    }
  for (int i = (0), i_end_ = (m); i < i_end_; ++i) assert(b[c[i]] == i);
  for (int i = (0), i_end_ = (n); i < i_end_; ++i) assert(c[b[i]] == a[i]);
  printf("%d\n", m);
  for (int i = (0), i_end_ = (n); i < i_end_; ++i) printf("%d ", b[i] + 1);
  printf("\n");
  for (int i = (0), i_end_ = (m); i < i_end_; ++i) printf("%d ", c[i] + 1);
  printf("\n");
  return 0;
}
