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 inf = 1e9 + 7;
const long long llinf = 1ll * inf * inf;
const double eps = 1e-12;
const int N = 1e5 + 10;
int f[N], h[N], g[N];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= int(n); ++i) scanf("%d", &f[i]);
  map<int, vector<int> > uf;
  for (int i = 1; i <= int(n); ++i) uf[f[i]].push_back(i);
  int m = uf.size();
  auto itr = uf.begin();
  for (int x = 1; x <= m; ++x) {
    h[x] = itr->first;
    vector<int> nv = itr->second;
    itr++;
    for (int y : nv) g[y] = x;
  }
  int bad = 0;
  for (int x = 1; x <= m; ++x)
    if (g[h[x]] != x) bad = 1;
  for (int x = 1; x <= n; ++x)
    if (h[g[x]] != f[x]) bad = 1;
  if (bad)
    puts("-1");
  else {
    printf("%d\n", m);
    for (int x = 1; x <= int(n); ++x) printf("%d%c", g[x], x == n ? '\n' : ' ');
    for (int x = 1; x <= int(m); ++x) printf("%d%c", h[x], x == m ? '\n' : ' ');
  }
  return 0;
}
