Problem: You are given an array of n positive integers a_1, a_2, ..., a_n. You can perform the following operation any number of times: select several distinct indices i_1, i_2, ..., i_k (1 ≤ i_j ≤ n) and move the number standing at the position i_1 to the position i_2, the number at the position i_2 to the position i_3, ..., the number at the position i_k to the position i_1. In other words, the operation cyclically shifts elements: i_1 → i_2 → … i_k → i_1.

For example, if you have n=4, an array a_1=10, a_2=20, a_3=30, a_4=40, and you choose three indices i_1=2, i_2=1, i_3=4, then the resulting array would become a_1=20, a_2=40, a_3=30, a_4=10.

Your goal is to make the array sorted in non-decreasing order with the minimum number of operations. The additional constraint is that the sum of cycle lengths over all operations should be less than or equal to a number s. If it's impossible to sort the array while satisfying that constraint, your solution should report that as well.

Input

The first line of the input contains two integers n and s (1 ≤ n ≤ 200 000, 0 ≤ s ≤ 200 000)—the number of elements in the array and the upper bound on the sum of cycle lengths.

The next line contains n integers a_1, a_2, ..., a_n—elements of the array (1 ≤ a_i ≤ 10^9).

Output

If it's impossible to sort the array using cycles of total length not exceeding s, print a single number "-1" (quotes for clarity).

Otherwise, print a single number q— the minimum number of operations required to sort the array.

On the next 2 ⋅ q lines print descriptions of operations in the order they are applied to the array. The description of i-th operation begins with a single line containing one integer k (1 ≤ k ≤ n)—the length of the cycle (that is, the number of selected indices). The next line should contain k distinct integers i_1, i_2, ..., i_k (1 ≤ i_j ≤ n)—the indices of the cycle.

The sum of lengths of these cycles should be less than or equal to s, and the array should be sorted after applying these q operations.

If there are several possible answers with the optimal q, print any of them.

Examples

Input

5 5
3 2 3 1 1


Output

1
5
1 4 2 3 5 


Input

4 3
2 1 4 3


Output

-1

Input

2 0
2 2


Output

0

Note

In the first example, it's also possible to sort the array with two operations of total length 5: first apply the cycle 1 → 4 → 1 (of length 2), then apply the cycle 2 → 3 → 5 → 2 (of length 3). However, it would be wrong answer as you're asked to use the minimal possible number of operations, which is 1 in that case.

In the second example, it's possible to the sort the array with two cycles of total length 4 (1 → 2 → 1 and 3 → 4 → 3). However, it's impossible to achieve the same using shorter cycles, which is required by s=3.

In the third example, the array is already sorted, so no operations are needed. Total length of empty set of cycles is considered to be zero.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int n, m, a[200010], b[200010], x[200010], y[200010], f[200010], r[200010],
    c[200010], p, d[200010], q;
map<int, int> h;
inline int fa(int i) { return f[i] == i ? i : f[i] = fa(f[i]); }
inline void unit(int i, int j) {
  i = fa(i);
  j = fa(j);
  if (i != j) f[i] = j;
}
int main() {
  int i, j, k;
  scanf("%d%d", &n, &m);
  for (i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    b[i] = a[i];
  }
  sort(b + 1, b + n + 1);
  for (i = 1; i <= n; i++)
    if (a[i] != b[i]) m--;
  if (m < 0) {
    printf("-1\n");
    return 0;
  }
  for (i = 1; i <= n; i++) f[i] = i;
  for (i = n; i > 0; i--)
    if (a[i] != b[i]) {
      r[i] = h[b[i]];
      h[b[i]] = i;
    }
  for (i = 1; i <= n; i++)
    if (a[i] != b[i]) {
      x[i] = h[a[i]];
      y[x[i]] = i;
      h[a[i]] = r[h[a[i]]];
      unit(i, x[i]);
    }
  for (i = 1, j = 0; i <= n; i++)
    if (a[i] != b[i]) {
      if (j && b[i] == b[j]) {
        if (fa(i) != fa(j)) {
          f[fa(i)] = fa(j);
          swap(y[i], y[j]);
          x[y[i]] = i;
          x[y[j]] = j;
        }
      }
      j = i;
    }
  for (i = 1; i <= n; i++)
    if (a[i] != b[i] && fa(i) == i) c[++p] = i;
  if (m >= 3 && p >= 3) {
    m = min(m, p);
    for (i = m; i > 0; i--) d[++q] = x[c[i]];
    for (i = 1; i < m; i++) {
      swap(x[c[i]], x[c[i + 1]]);
      f[fa(c[i])] = fa(c[i + 1]);
    }
  }
  p = 0;
  for (i = 1; i <= n; i++)
    if (a[i] != b[i] && fa(i) == i) c[++p] = i;
  printf("%d\n", p + (q > 0));
  for (i = 1; i <= n; i++)
    if (a[i] != b[i] && fa(i) == i) {
      for (j = x[i], k = 1; j != i; j = x[j]) k++;
      printf("%d\n%d", k, i);
      for (j = x[i], k = 1; j != i; j = x[j]) printf(" %d", j);
      printf("\n");
    }
  if (q) {
    printf("%d\n", q);
    for (i = 1; i <= q; i++) printf("%d ", d[i]);
    printf("\n");
  }
  return 0;
}