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;
const int maxn = 100010;
int f[maxn];
int ansn[maxn], ansm[maxn];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  ;
  int n;
  cin >> n;
  int num = 0;
  for (int j = 1; j <= n; ++j) {
    cin >> f[j];
    if (f[j] == j) {
      ++num;
      ansm[num] = j;
      ansn[j] = num;
    }
  }
  bool flag = 1;
  for (int j = 1; j <= n; ++j) {
    if (f[j] != j) {
      int zx = f[j];
      if (f[zx] != zx)
        flag = 0;
      else {
        ansn[j] = ansn[zx];
      }
    }
  }
  if (flag == 0)
    cout << "-1" << endl;
  else {
    cout << num << endl;
    for (int j = 1; j <= n; ++j) cout << ansn[j] << " ";
    cout << endl;
    for (int j = 1; j <= num; ++j) cout << ansm[j] << " ";
    cout << endl;
  }
  return 0;
}
