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 f[1000010], g[1000010], h[1000010];
int buf[1000010];
int th[1000010];
int n, m;
int main() {
  while (cin >> n) {
    memset(buf, 0, sizeof(buf));
    m = 0;
    int a = 0;
    for (int i = 1; i <= n; i++) {
      scanf("%d", &f[i]);
      if (buf[f[i]] == 0) {
        buf[f[i]] = 1;
        m++;
      }
      if (i == f[i]) {
        h[++a] = f[i];
        g[i] = a;
        th[f[i]] = a;
      }
    }
    if (m > a) {
      cout << -1 << endl;
      continue;
    }
    cout << m << endl;
    for (int i = 1; i <= m; i++) {
      g[h[i]] = i;
      int a = h[i];
      g[f[a]] = i;
    }
    for (int i = 1; i <= n; i++) {
      g[i] = th[f[i]];
    }
    for (int i = 1; i < n; i++) {
      printf("%d ", g[i]);
    }
    printf("%d\n", g[n]);
    for (int i = 1; i < m; i++) {
      printf("%d ", h[i]);
    }
    printf("%d\n", h[m]);
  }
  return 0;
}
