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 long long inf = (long long)1e+9 + 7ll;
const long long linf = (long long)1e+18 + 7ll;
const long double eps = (long double)1e-10;
const long double pi = acosl((long double)-1.0);
const int alph = 26;
static char buff[(int)2e6 + 17];
const int maxn = (int)1e6 + 17;
int n;
int f[maxn], p[maxn], g[maxn];
bool read() {
  if (scanf("%d", &n) != 1) return false;
  for (int i = 0; i < n; ++i) scanf("%d", &f[i]), --f[i];
  return true;
}
void solve() {
  vector<int> h;
  for (int i = 0; i < n; ++i) h.push_back(f[i]);
  sort(h.begin(), h.end());
  h.resize(unique(h.begin(), h.end()) - h.begin());
  for (int i = 0; i < ((int)(h.size())); ++i) p[h[i]] = i;
  for (int i = 0; i < n; ++i) g[i] = p[f[i]];
  for (int i = 0; i < ((int)(h.size())); ++i)
    if (g[h[i]] != i) {
      printf("-1\n");
      return;
    }
  printf("%d\n", ((int)(h.size())));
  for (long long i = 0ll; i < (long long)n; ++i) printf("%d ", g[i] + 1);
  printf("\n");
  for (long long i = 0ll; i < (long long)((int)(h.size())); ++i)
    printf("%d ", h[i] + 1);
}
int main() {
  while (read()) solve();
  return 0;
}
