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;
template <class T>
inline T mod_v(T num) {
  if (num >= 0)
    return num % 1000000007;
  else
    return (num % 1000000007 + 1000000007) % 1000000007;
}
template <class T>
inline T gcd(T a, T b) {
  a = abs(a);
  b = abs(b);
  while (b) b ^= a ^= b ^= a %= b;
  return a;
}
template <class T>
T fast_pow(T n, T p) {
  if (p == 0) return 1;
  if (p % 2) {
    T g = mod_v(mod_v(n) * mod_v(fast_pow(n, p - 1)));
    return g;
  } else {
    T g = fast_pow(n, p / 2);
    g = mod_v(mod_v(g) * mod_v(g));
    return g;
  }
}
template <class T>
inline T modInverse(T n) {
  return fast_pow(n, 1000000007 - 2);
}
bool equalTo(double a, double b) {
  if (fabs(a - b) <= 1e-9)
    return true;
  else
    return false;
}
bool notEqual(double a, double b) {
  if (fabs(a - b) > 1e-9)
    return true;
  else
    return false;
}
bool lessThan(double a, double b) {
  if (a + 1e-9 < b)
    return true;
  else
    return false;
}
bool lessThanEqual(double a, double b) {
  if (a < b + 1e-9)
    return true;
  else
    return false;
}
bool greaterThan(double a, double b) {
  if (a > b + 1e-9)
    return true;
  else
    return false;
}
bool greaterThanEqual(double a, double b) {
  if (a + 1e-9 > b)
    return true;
  else
    return false;
}
template <class T>
inline int in(register T& num) {
  register char c = 0;
  num = 0;
  bool n = false;
  while (c < 33) c = getchar();
  while (c > 33) {
    if (c == '-')
      n = true;
    else
      num = num * 10 + c - '0';
    c = getchar();
  }
  num = n ? -num : num;
  return 1;
}
vector<int> lst[100005];
int n, m;
int x[100005], y[100005];
int main() {
  cin >> n;
  int val;
  for (int i = 1; i <= n; i++) {
    in(val);
    lst[val].push_back(i);
  }
  bool ok = true;
  for (int i = 1; i <= n && ok; i++) {
    if (((int)lst[i].size())) {
      int p = lower_bound((lst[i]).begin(), (lst[i]).end(), i) - lst[i].begin();
      if (lst[i][p] != i) {
        ok = false;
      } else {
        y[++m] = lst[i][p];
        for (int j = 0; j < ((int)lst[i].size()); j++) x[lst[i][j]] = m;
      }
    }
  }
  if (!ok)
    printf("-1");
  else {
    printf("%d\n", m);
    for (int i = 1; i <= n; i++) printf("%d ", x[i]);
    printf("\n");
    for (int i = 1; i <= m; i++) printf("%d ", y[i]);
  }
  return 0;
}