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 = 1e5 + 10;
int n;
int f[MAXN];
bool read() {
  if (scanf("%d", &n) < 1) {
    return 0;
  }
  for (int i = 0; i < int(n); ++i) {
    scanf("%d", &f[i]);
    --f[i];
  }
  return 1;
}
vector<int> inv_f[MAXN];
int g[MAXN];
int h[MAXN];
bool solve() {
  for (int i = 0; i < int(n); ++i) {
    inv_f[i].clear();
  }
  for (int i = 0; i < int(n); ++i) {
    inv_f[f[i]].push_back(i);
  }
  memset(g, -1, sizeof g);
  int m = 0;
  for (int i = 0; i < int(n); ++i) {
    if (inv_f[i].empty()) {
      continue;
    }
    if (f[i] != i) {
      return 0;
    }
    for (int j : inv_f[i]) {
      g[j] = m;
    }
    h[m] = i;
    ++m;
  }
  for (int x = 0; x < int(m); ++x) {
    assert(g[h[x]] == x);
  }
  for (int x = 0; x < int(n); ++x) {
    assert(h[g[x]] == f[x]);
  }
  printf("%d\n", m);
  for (int i = 0; i < int(n); ++i) {
    printf("%d ", g[i] + 1);
  }
  puts("");
  for (int i = 0; i < int(m); ++i) {
    printf("%d ", h[i] + 1);
  }
  puts("");
  return 1;
}
int main() {
  while (read()) {
    if (!solve()) {
      puts("-1");
    }
  }
  return 0;
}
