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;
int const siz = 1e9, bound = 5e5, mod = siz + 9;
long long const oo = 1e18;
int f[bound], h[bound], g[bound];
int n;
vector<int> nxt[bound];
set<int> st;
int main() {
  memset(h, -1, sizeof h);
  memset(g, -1, sizeof g);
  scanf("%d", &n);
  for (int(i) = 1; (i) <= (n); (i)++) {
    scanf("%d", &f[i]);
    nxt[f[i]].push_back(i);
    st.insert(f[i]);
  }
  if (st.size() == 1) {
    printf("%d\n", 1);
    for (int(i) = 1; (i) <= (n); (i)++) printf("%d ", 1);
    puts("");
    printf("%d\n", *st.begin());
  } else {
    int m = 0;
    for (int(i) = 1; (i) <= (n); (i)++) {
      if (!nxt[i].empty()) {
        m++;
        h[m] = i;
      }
      for (int(j) = 0; (j) < (nxt[i].size()); (j)++) g[nxt[i][j]] = m;
    }
    for (int(i) = 1; (i) <= (n); (i)++) {
      if (h[g[i]] != f[i]) {
        puts("-1");
        return 0;
      }
    }
    for (int(i) = 1; (i) <= (m); (i)++) {
      if (g[h[i]] != i) {
        puts("-1");
        return 0;
      }
    }
    printf("%d\n", m);
    for (int(i) = 1; (i) <= (n); (i)++) printf("%d ", g[i]);
    puts("");
    for (int(i) = 1; (i) <= (m); (i)++) printf("%d ", h[i]);
    puts("");
  }
  return 0;
}
