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;
struct node {
  int n;
  int at;
};
bool operator<(node a, node b) { return a.at > b.at; }
int ds[100005], n, m = 0;
bool valid = true;
void preprocess() {}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  vector<int> h(n + 1, 0);
  for (int i = 0; i < n; i++) cin >> ds[i + 1];
  for (int i = 0; i < n; i++) {
    if (ds[ds[i + 1]] != ds[i + 1]) {
      valid = false;
      break;
    } else if (ds[i + 1] == i + 1)
      h[i + 1] = ++m;
  }
  if (valid) {
    cout << m << "\n";
    for (int i = 0; i < n; i++) cout << h[ds[i + 1]] << " ";
    cout << "\n";
    for (int i = 0; i < n; i++)
      if (h[i + 1] > 0) cout << ds[ds[i + 1]] << " ";
    cout << "\n";
  } else
    cout << -1;
  return 0;
}
