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[200010], b[200010], nxt[200010], cnt, n, s;
void solve(vector<int> &v, int i, int &j) {
  while (a[j] == a[i]) j++;
  if (j >= n) return;
  if (b[j] == a[i]) {
    swap(a[i], a[j]);
    cnt++;
    v.push_back(j++);
    solve(v, i, nxt[lower_bound(b, b + n, a[i]) - b]);
  }
}
int main() {
  cin >> n >> s;
  for (int i = 0; i < n; i++) {
    scanf("%d", a + i);
    b[i] = a[i];
    nxt[i] = i;
  }
  sort(b, b + n);
  a[n] = b[n] = -1;
  cnt = 0;
  vector<vector<int> > ans;
  for (int i = 0; i < n; i++) {
    if (b[i] == a[i]) continue;
    ans.push_back(vector<int>(1, i));
    cnt++;
    nxt[lower_bound(b, b + n, b[i]) - b] = i + 1;
    auto &v = ans.back();
    solve(v, i, nxt[lower_bound(b, b + n, a[i]) - b]);
    for (int k = 0; k < v.size(); k++) {
      int j0 = v[k];
      int &j = nxt[lower_bound(b, b + n, b[j0]) - b];
      while (b[j0] == b[j]) {
        if (b[j] != a[j]) {
          cnt++;
          vector<int> v1(1, j);
          int j1 = j++;
          solve(v1, j1, nxt[lower_bound(b, b + n, a[j1]) - b]);
          v.insert(v.begin() + k, v1.begin(), v1.end());
        } else
          j++;
      }
    }
  }
  if (cnt > s) {
    printf("-1\n");
  } else if (ans.size() < 3 || s - cnt < 3) {
    printf("%d\n", ans.size());
    while (ans.size() > 0) {
      auto &v = ans.back();
      printf("%d\n", v.size());
      for (int i : v) {
        printf("%d ", i + 1);
      }
      printf("\n");
      ans.pop_back();
    }
  } else {
    int x = max((int)ans.size() - s + cnt, 0), size = 0;
    for (int i = x; i < ans.size(); i++) {
      size += ans[i].size();
    }
    printf("%d\n%d\n", x + 2, size);
    for (int i = x; i < ans.size(); i++) {
      auto &v = ans[i];
      for (int i : v) {
        printf("%d ", i + 1);
      }
    }
    printf("\n%d\n", ans.size() - x);
    for (int i = ans.size() - 1; i >= x; i--) {
      printf("%d ", ans[i][0] + 1);
    }
    printf("\n");
    ans.resize(x);
    while (ans.size() > 0) {
      auto &v = ans.back();
      printf("%d\n", v.size());
      for (int i : v) {
        printf("%d ", i + 1);
      }
      printf("\n");
      ans.pop_back();
    }
  }
  return 0;
}