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 f[1000001], g[1000001];
int h[1000001], hcount;
int isUsed[1000001];
int isTrue;
int kinds;
int main() {
  int n;
  int t;
  while (scanf("%d", &n) != EOF) {
    memset(f, 0, sizeof(f));
    memset(g, 0, sizeof(g));
    memset(isUsed, 0, sizeof(isUsed));
    isTrue = 1;
    hcount = 0, kinds = 1;
    for (t = 1; t <= n; t++) {
      int in;
      scanf("%d", &in);
      if (in <= n && isTrue) {
        if (!isUsed[in]) {
          isUsed[in] = kinds++;
          g[t] = isUsed[in];
          f[t] = in;
          h[hcount++] = in;
        } else {
          g[t] = isUsed[in];
          f[t] = in;
        }
      } else {
      }
    }
    for (t = 0; t < hcount; t++) {
      if (!(h[t] == f[h[t]])) {
        isTrue = 0;
      }
    }
    if (isTrue) {
      printf("%d\n", hcount);
      for (t = 1; t <= n; t++) {
        printf("%d ", g[t]);
      }
      printf("\n");
      for (t = 0; t < hcount; t++) {
        printf("%d ", h[t]);
      }
      printf("\n");
    } else {
      printf("-1\n");
    }
  }
  return 0;
}
