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;
long long r, m, n, a[2000005], b[2000005], c[2000005], xz, yz;
vector<long long> v;
map<long long, long long> co;
int main() {
  ios_base::sync_with_stdio(0);
  long long k = 0, t = 0, x, sum = 0, q, y = 0, l = 0, i, j, z, p;
  cin >> n;
  for (i = 1; i <= n; i++) {
    cin >> a[i];
    if (co[a[i]] == 0) {
      co[a[i]] = 1;
      v.push_back(a[i]);
    }
  }
  m = v.size();
  sort(v.begin(), v.end());
  for (i = 0; i < m; i++) {
    t = v[i];
    co[t] = i + 1;
  }
  for (i = 1; i <= n; i++) {
    t = a[i];
    b[i] = co[t];
  }
  for (i = 1; i <= m; i++) {
    c[i] = v[i - 1];
  }
  x = 0;
  for (i = 1; i <= m; i++) {
    if (b[c[i]] != i) {
      x = 1;
    }
    if (x) break;
  }
  for (i = 1; i <= n; i++) {
    if (c[b[i]] != a[i]) {
      x = 1;
    }
    if (x) break;
  }
  if (x)
    cout << "-1";
  else {
    cout << m << endl;
    for (i = 1; i <= n; i++) cout << b[i] << " ";
    cout << endl;
    for (i = 1; i <= m; i++) cout << c[i] << " ";
  }
  return 0;
}
