Problem: 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int n;
int a[100005], h[100005], g[100005];
int m;
map<int, vector<int> > mp;
map<int, vector<int> >::iterator it;
int main() {
  int t, i;
  t = 1;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%d", a + i);
    mp[a[i]].push_back(i);
  }
  m = 1;
  for (it = mp.begin(); it != mp.end(); it++) {
    for (i = 0; i < (int)it->second.size(); i++) {
      if (g[it->second[i]] != 0) {
        printf("-1");
        return 0;
      }
      g[(it->second)[i]] = m;
    }
    h[m++] = it->first;
  }
  m--;
  for (i = 1; i <= n; i++)
    if (g[i] == 0) {
      assert(0);
    }
  for (i = 1; i <= n; i++)
    if (h[g[i]] != a[i]) {
      printf("-1");
      return 0;
    }
  for (i = 1; i <= m; i++)
    if (g[h[i]] != i) {
      printf("-1");
      return 0;
    }
  printf("%d\n%d", m, g[1]);
  for (i = 2; i <= n; i++) printf(" %d", g[i]);
  printf("\n%d", h[1]);
  for (i = 2; i <= m; i++) printf(" %d", h[i]);
  return 0;
}