Problem: 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 50, maxm = 1e5 + 50;
const int inf = 0x3f3f3f3f, mod = (int)1e9 + 7;
const long long INF = 0x3f3f3f3f3f3f3f3f;
typedef int arrn[maxn];
typedef long long arrm[maxn];
template <class Q>
void gint(Q &ret, int sig = 1, char ch = getchar()) {
  while (!isdigit(ch) && ch != '-') ch = getchar();
  if (ch == '-') sig = -1, ch = getchar();
  for (ret = 0; isdigit(ch); ch = getchar()) ret = ret * 10 + ch - '0';
  ret *= sig;
}
int n, m, sz;
arrn f, g, h, v, pa;
int find(int x) { return x == pa[x] ? x : pa[x] = find(pa[x]); }
int main() {
  cin >> n;
  for (int i = 1; i <= n; ++i) cin >> f[i], pa[i] = i;
  for (int i = 1; i <= n; ++i) pa[find(i)] = pa[find(f[i])];
  for (int i = 1; i <= n; ++i)
    if (i == find(i)) g[i] = ++m, h[m] = f[i];
  for (int i = 1; i <= n; ++i)
    if (i != find(i)) {
      g[i] = g[find(i)];
      if (f[i] != f[find(i)]) puts("-1"), exit(0);
    }
  cout << m << endl;
  for (int i = 1; i <= n; ++i) cout << g[i] << " ";
  cout << endl;
  for (int i = 1; i <= m; ++i) cout << h[i] << " ";
}