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, p[maxn], f[maxn], g[maxn], h[maxn];
int find(int x) { return p[x] == x ? x : p[x] = find(p[x]); }
vector<int> G[maxn];
int main(void) {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) p[i] = i;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &f[i]);
  }
  for (int i = 1; i <= n; i++) {
    int u = f[i], v = i;
    int fu = find(u);
    int fv = find(v);
    if (fu == fv) continue;
    if (fu < fv) {
      G[fu].push_back(v);
      p[fv] = fu;
    } else {
      G[fv].push_back(u);
      p[fu] = fv;
    }
  }
  int m = 0;
  for (int i = 1; i <= n; i++) {
    if (p[i] == i) {
      g[i] = ++m;
      for (int k = 0; k < G[i].size(); k++) {
        g[G[i][k]] = m;
      }
    }
  }
  int flag = 1;
  for (int i = 1; i <= n; i++) {
    if (h[g[i]] == 0)
      h[g[i]] = f[i];
    else {
      if (f[i] != h[g[i]]) {
        flag = 0;
        break;
      }
    }
  }
  if (flag == 0)
    printf("-1\n");
  else {
    printf("%d\n", m);
    for (int i = 1; i <= n; i++) {
      if (i == 1)
        printf("%d", g[i]);
      else
        printf(" %d", g[i]);
    }
    printf("\n");
    for (int i = 1; i <= m; i++) {
      if (i == 1)
        printf("%d", h[i]);
      else
        printf(" %d", h[i]);
    }
  }
  return 0;
}
