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 = (int)1e6 + 22;
const int ALF = 26;
void cry() {
  cout << -1;
  exit(0);
}
int n, m;
int f[MAXN], g[MAXN], h[MAXN];
bool check() {
  for (int i = 0; i < m; i++) {
    if (g[h[i]] != i) return false;
  }
  for (int i = 0; i < n; i++) {
    if (h[g[i]] != f[i]) return false;
  }
  return true;
}
void solve() {
  cin >> n;
  vector<long long> vv;
  for (int i = 0; i < n; i++) {
    cin >> f[i];
    vv.push_back(--f[i]);
  }
  sort(vv.begin(), vv.end());
  vv.resize(unique(vv.begin(), vv.end()) - vv.begin());
  m = (int)vv.size();
  for (int i = 0; i < m; i++) h[i] = vv[i];
  for (int i = 0; i < n; i++)
    g[i] = lower_bound(vv.begin(), vv.end(), f[i]) - vv.begin();
  if (!check()) cry();
  cout << m << "\n";
  for (int i = 0; i < n; i++) cout << g[i] + 1 << " ";
  cout << "\n";
  for (int i = 0; i < m; i++) cout << h[i] + 1 << " ";
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  solve();
  return 0;
}
