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 le[200010], ev[200010], ecnt, pe[200010];
void addEdge(int u, int v) {
  ecnt++;
  pe[ecnt] = le[u];
  le[u] = ecnt;
  ev[ecnt] = v;
}
int vis[200010], stable, vise[200010];
int head[200010], pa[200010];
int findset(int u) {
  if (u != pa[u]) pa[u] = findset(pa[u]);
  return pa[u];
}
vector<int> h;
int g[200010], n, f[200010];
vector<int> st[200010];
int main() {
  scanf("%d", &n);
  for (int i = 1, ed_i = n; i <= ed_i; i++) pa[i] = i;
  for (int i = 1, ed_i = n; i <= ed_i; i++) {
    scanf("%d", &f[i]);
    int x = findset(i), y = findset(f[i]);
    if (x != y) pa[pa[x]] = pa[y];
  }
  for (int i = 1, ed_i = n; i <= ed_i; i++) {
    if (f[i] != f[f[i]]) {
      puts("-1");
      return 0;
    }
    st[findset(i)].push_back(i);
    if (f[i] == i) {
      head[findset(i)] = i;
    }
  }
  for (int i = 1, ed_i = n; i <= ed_i; i++) {
    if (!vis[findset(i)]) {
      if (!head[pa[i]]) {
        puts("-1");
        return 0;
      }
      vis[pa[i]] = 1;
      h.push_back(head[findset(i)]);
      for (int u : st[findset(i)]) {
        g[u] = h.size();
      }
    }
  }
  printf("%d\n", h.size());
  for (int i = 1, ed_i = n; i <= ed_i; i++) {
    printf("%d ", g[i]);
  }
  puts("");
  for (auto u : h) {
    printf("%d ", u);
  }
  return 0;
}
