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 a[200000], b[200000];
vector<int> pos[200000];
int p[200000], nxt[200000];
int parent[200000];
int find(int n) {
  if (parent[n] != n) parent[n] = find(parent[n]);
  return parent[n];
}
int visited[200000];
vector<int> cycles[200000];
int main() {
  int i;
  int n, s;
  scanf("%d %d", &n, &s);
  for (i = 0; i < n; i++) scanf("%d", &a[i]), b[i] = a[i];
  sort(b, b + n);
  for (i = 0; i < n; i++) a[i] = lower_bound(b, b + n, a[i]) - b;
  for (i = 0; i < n; i++) b[i] = a[i];
  sort(b, b + n);
  int j, c = 0;
  for (i = 0; i < n; i++) {
    if (a[i] != b[i]) pos[b[i]].push_back(i), c++;
    parent[i] = i, p[i] = -1;
  }
  if (c > s) {
    printf("-1\n");
    return 0;
  }
  for (i = 0; i < n; i++) {
    if (a[i] == b[i])
      nxt[i] = i;
    else
      nxt[i] = pos[a[i]].back(), pos[a[i]].pop_back();
    int pa = find(i), push_back = find(nxt[i]);
    if (pa != push_back) parent[pa] = push_back;
  }
  for (i = 0; i < n; i++) {
    if (a[i] == b[i]) continue;
    if (p[a[i]] != -1) {
      int pa = find(p[a[i]]), push_back = find(i);
      if (pa != push_back) {
        int t = nxt[p[a[i]]];
        nxt[p[a[i]]] = nxt[i];
        nxt[i] = t;
        parent[pa] = push_back;
      }
    }
    p[a[i]] = i;
  }
  int cc = 0;
  for (i = 0; i < n; i++) {
    if (!visited[i] && (i != nxt[i])) {
      int u = i;
      do cycles[cc].push_back(u), visited[u] = 1, u = nxt[u];
      while (u != i);
      cc++;
    }
  }
  int x = min(s - c, cc);
  printf("%d\n", cc - x + min(x, 2));
  int sum = 0;
  if (x > 0) {
    for (i = 0; i < x; i++) sum += cycles[i].size();
    printf("%d\n", sum);
    for (i = 0; i < x; i++) {
      for (j = 0; j < cycles[i].size(); j++) printf("%d ", cycles[i][j] + 1);
    }
    printf("\n");
    if (x > 1) {
      printf("%d\n", x);
      for (i = x - 1; i >= 0; i--) printf("%d ", cycles[i][0] + 1);
      printf("\n");
    }
  }
  for (i = x; i < cc; i++) {
    printf("%d\n", cycles[i].size());
    for (j = 0; j < cycles[i].size(); j++) printf("%d ", cycles[i][j] + 1);
    printf("\n");
  }
  return 0;
}