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;
bool visited[1111111];
int n, m, f[1111111], g[1111111], h[1111111];
bool check() {
  if (n == 1 && f[1] != 1) return false;
  for (int x = 1; x <= m; x++) {
    if (g[h[x]] != x) return false;
  }
  for (int x = 1; x <= n; x++) {
    if (h[g[x]] != f[x]) return false;
  }
  return true;
}
void compress(int n, int* in, int* out) {
  map<int, int> mp;
  for (int i = 1; i <= n; i++)
    out[i] = mp.emplace(in[i], mp.size()).first->second + 1;
}
int main() {
  int i, j, k, l;
  while (scanf("%d", &n) != EOF) {
    for (i = 1; i <= n; i++) scanf("%d", &f[i]);
    compress(n, f, g);
    memset(visited, 0, sizeof(visited));
    for (i = 1; i <= n; i++) {
      if (!visited[g[i]]) {
        visited[g[i]] = true;
        h[g[i]] = f[i];
      }
    }
    for (m = 0, i = 1; i <= n; i++) m = max(m, g[i]);
    if (!check())
      puts("-1");
    else {
      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]);
      puts("");
    }
  }
  return 0;
}
