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;
long long inf = 2147483647;
const long long ll = 9223372036854775807, ninf = 1000000000;
const double eps = 1e-6;
const long long nll = 1000000000000000000;
class MulticoreProcessingEasy {
 public:
  int fastestTime(int jobLength, int corePenalty, vector<int> speed,
                  vector<int> cores) {}
};
int main() {
  int a;
  while (cin >> a) {
    int A[100005] = {};
    for (int i = 1; i <= a; i++) cin >> A[i];
    int no = 0;
    for (int i = 1; i <= a; i++) {
      if (A[i] != A[A[i]]) {
        no = 1;
      }
    }
    if (no) {
      cout << -1 << '\n';
      continue;
    }
    int b[100005] = {};
    int c[100005] = {};
    int vis[100005] = {};
    int now = 0;
    for (int i = 1; i <= a; i++) {
      if (!vis[A[i]]) {
        now++;
        vis[A[i]] = now;
        c[now] = A[i];
      }
      b[i] = vis[A[i]];
    }
    cout << now << '\n';
    for (int i = 1; i <= a; i++) cout << b[i] << " ";
    cout << '\n';
    for (int i = 1; i <= now; i++) cout << c[i] << " ";
    cout << '\n';
  }
  return 0;
}