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 + 5;
const int MOD = 1e9 + 7;
const int INF = 2e9;
int arr[MAXN], G[MAXN], H[MAXN];
vector<int> mem[MAXN];
int main() {
  int N = 0, M = 0, timer = 1;
  scanf("%i", &N);
  for (int i = 1; i <= N; i++) {
    scanf("%i", &arr[i]);
    mem[arr[i]].push_back(i);
  }
  for (int i = 1; i <= N; i++)
    if (!G[arr[i]]) G[arr[i]] = timer++;
  for (int i = 1; i <= N; i++) {
    M = max(M, G[i]);
    H[G[i]] = arr[i];
  }
  for (int i = 1; i <= M; i++) {
    if (!H[i]) return printf("-1"), 0;
    for (int j = 0; j < mem[H[i]].size(); j++) G[mem[H[i]][j]] = i;
  }
  for (int i = 1; i <= N; i++) {
    if (!G[i]) return printf("-1"), 0;
    if (H[G[i]] != arr[i]) return printf("-1"), 0;
  }
  for (int i = 1; i <= M; i++)
    if (G[H[i]] != i) return printf("-1"), 0;
  printf("%i\n", M);
  for (int i = 1; i <= N; i++) printf("%i ", G[i]);
  printf("\n");
  for (int i = 1; i <= M; i++) printf("%i ", H[i]);
  return 0;
}
