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>
int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); }
int lcm(int a, int b) { return ((a * b) / gcd(a, b)); }
bool isPrime(long long n) {
  if (n <= 1) return 0;
  if (n <= 3) return 1;
  if (n % 2 == 0)
    return 0;
  else {
    for (long long i = 5; i * i <= n; i += 2) {
      if (n % i == 0) return 0;
    }
    return 1;
  }
}
template <class T>
inline T bigmod(T p, T e, T M) {
  long long ret = 1;
  for (; e > 0; e >>= 1) {
    if (e & 1) ret = (ret * p) % M;
    p = (p * p) % M;
  }
  return (T)ret;
}
template <class T>
inline T gcd(T a, T b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
template <class T>
inline T modinverse(T a, T M) {
  return bigmod(a, M - 2, M);
}
using namespace std;
vector<long long> h;
long long g[100005], f[100005];
int main() {
  long long i, j, k, l, m, n;
  cin >> n;
  h.push_back(0);
  for (i = 1; i <= n; i++) {
    cin >> f[i];
    if (f[i] == i) {
      h.push_back(i);
      g[h[h.size() - 1]] = h.size() - 1;
    }
  }
  for (i = 1; i <= n; i++) {
    if (g[f[i]] == 0) {
      cout << -1;
      return 0;
    }
  }
  cout << h.size() - 1 << '\n';
  for (i = 1; i <= n; i++) {
    cout << g[f[i]] << " ";
  }
  cout << '\n';
  for (i = 1; i < h.size(); i++) {
    cout << h[i] << " ";
  }
}
