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;
template <class T>
T abs(T a) {
  return a >= 0 ? a : -a;
}
template <class T>
T gcd(T a, T b) {
  return b ? gcd(b, a % b) : a;
}
template <class T>
T mod(T a, T b) {
  T ret = a % b;
  if (ret < 0) ret += b;
  return ret;
}
long long mulmod(long long a, long long b, long long c) {
  if (b == 0LL) return 0LL;
  long long ret = mulmod(a, b >> 1, c);
  ret = (ret + ret) % c;
  if (b & 1LL) ret = (ret + a) % c;
  return ret;
}
long long powmod(long long a, long long b, long long c) {
  if (b == 0LL) return 1LL;
  long long ret = powmod(a, b >> 1, c);
  ret = ret * ret % c;
  if (b & 1LL) ret = ret * a % c;
  return ret;
}
template <class T>
inline void wt(T x) {
  if (x < 0) putchar('-'), x = -x;
  if (x < 10)
    putchar(x + '0');
  else
    wt(x / 10), putchar(x % 10 + '0');
}
template <class T>
inline void rd(T& x) {
  bool f = 0;
  char ch;
  for (ch = getchar(); ch <= 32; ch = getchar())
    ;
  if (ch == '-') f = 1, ch = getchar();
  for (x = 0; ch > 32; ch = getchar()) x = (x << 3) + (x << 1) + ch - '0';
  if (f) x = -x;
}
const int N = 100005;
int tt;
int u, v, w, n, m;
int f[N], g[N], h[N];
int main() {
  rd(n);
  for (register int i = 1; i <= n; ++i) rd(f[i]);
  for (register int i = 1; i <= n; ++i)
    if (f[i] == i) {
      m++;
      h[m] = i;
      g[i] = m;
    }
  for (register int i = 1; i <= n; ++i)
    if (g[f[i]] == 0) {
      puts("-1");
      return 0;
    }
  wt(m);
  puts("");
  for (register int i = 1; i <= n; ++i) {
    if (i - 1) putchar(' ');
    wt(g[f[i]]);
  }
  puts("");
  for (register int i = 1; i <= m; ++i) {
    if (i - 1) putchar(' ');
    wt(h[i]);
  }
  puts("");
  return 0;
}
