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 arr[111111], a[111111], b[111111], n;
bool w1[111111], w2[111111];
int main() {
  while (cin >> n) {
    memset(a, 0, sizeof(a));
    memset(b, 0, sizeof(b));
    memset(w1, 0, sizeof(w1));
    memset(w2, 0, sizeof(w2));
    int cnt = 0;
    int ww = 0;
    for (int i = 1; i <= n; i++) {
      scanf("%d", &arr[i]);
      if (w1[arr[i]] == 0) cnt++, w1[arr[i]] = 1;
      if (arr[i] == i) ww++;
    }
    map<int, int> mp;
    if (ww == cnt) {
      cout << cnt << endl;
      int t = 0;
      for (int i = 1; i <= n; i++) {
        if (arr[i] == i) {
          w2[arr[i]] = 1;
          b[++t] = i;
          a[i] = t;
          mp[i] = t;
        }
      }
      for (int i = 1; i <= n; i++)
        if (a[i] == 0) {
          if (w2[arr[i]]) {
            a[i] = mp[arr[i]];
            continue;
          } else {
            w2[arr[i]] = 1;
            b[++t] = arr[i];
            a[i] = t;
          }
        }
      for (int i = 1; i <= n; i++) printf("%d%c", a[i], i == n ? '\n' : ' ');
      for (int i = 1; i <= cnt; i++)
        printf("%d%c", b[i], i == cnt ? '\n' : ' ');
    } else
      puts("-1");
  }
  return 0;
}
