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 m, f[111111], g[111111], h[111111], pos[111111];
int main() {
  while (~scanf("%d", &m)) {
    memset(g, -1, sizeof(g));
    memset(pos, -1, sizeof(pos));
    int n = 0;
    for (int i = 1; i <= m; i++) {
      scanf("%d", &f[i]);
      if (f[i] == i) h[++n] = i, g[i] = n, pos[i] = i;
    }
    if (!n)
      printf("-1\n");
    else {
      int gg = 0;
      for (int i = 1; i <= m; i++) {
        if (g[i] == -1) {
          if (pos[f[i]] == -1)
            gg = 1;
          else
            g[i] = g[pos[f[i]]];
        }
      }
      if (gg)
        printf("-1\n");
      else {
        printf("%d\n", n);
        for (int i = 1; i <= m; i++) printf("%d%c", g[i], i == m ? '\n' : ' ');
        for (int i = 1; i <= n; i++) printf("%d%c", h[i], i == n ? '\n' : ' ');
      }
    }
  }
  return 0;
}
