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;
inline long long rd() {
  long long x = 0;
  int ch = getchar(), f = 1;
  while (!isdigit(ch) && (ch != '-') && (ch != EOF)) ch = getchar();
  if (ch == '-') {
    f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = (x << 1) + (x << 3) + ch - '0';
    ch = getchar();
  }
  return x * f;
}
inline void rt(long long x) {
  if (x < 0) putchar('-'), x = -x;
  if (x >= 10)
    rt(x / 10), putchar(x % 10 + '0');
  else
    putchar(x + '0');
}
const int maxn = 100005;
int n, f[maxn], h[maxn], m, cnt[maxn], g[maxn];
int main() {
  n = rd();
  for (int i = 1; i <= (int)n; i++) f[i] = rd();
  for (int i = 1; i <= (int)n; i++) cnt[f[i]]++;
  for (int i = 1; i <= (int)n; i++)
    if (cnt[i] && f[i] != i) return puts("-1"), 0;
  for (int i = 1; i <= (int)n; i++)
    if (cnt[i]) h[++m] = i, g[i] = m;
  for (int i = 1; i <= (int)n; i++)
    if (f[i] != i) g[i] = g[f[i]];
  rt(m), putchar('\n');
  for (int i = 1; i <= (int)n; i++) rt(g[i]), putchar(' ');
  puts("");
  for (int i = 1; i <= (int)m; i++) rt(h[i]), putchar(' ');
  puts("");
}
