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;
const int MAXN = 200100;
int arr[MAXN], tmp[MAXN];
vector<int> ve[MAXN];
map<pair<int, int>, vector<int> > mapa;
int ind[MAXN];
vector<int> out[MAXN];
int sol = 0;
vector<int> t1, t2;
vector<int> comp;
map<int, int> aaa;
void tour(int x, int i) {
  while (ind[x] < ve[x].size()) {
    int y = ve[x][ind[x]++];
    int ind = mapa[{x, y}].back();
    mapa[{x, y}].pop_back();
    tour(y, ind);
  }
  out[sol].push_back(i);
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  int n, s;
  cin >> n >> s;
  for (int i = (0); i < (n); i++) {
    cin >> arr[i];
    comp.push_back(arr[i]);
  }
  sort(comp.begin(), comp.end());
  comp.resize(unique(comp.begin(), comp.end()) - comp.begin());
  for (int i = (0); i < (comp.size()); i++) {
    aaa[comp[i]] = i;
  }
  for (int i = (0); i < (n); i++) {
    arr[i] = aaa[arr[i]];
    tmp[i] = arr[i];
  }
  sort(tmp, tmp + n);
  for (int i = (0); i < (n); i++) {
    if (arr[i] != tmp[i]) {
      mapa[{tmp[i], arr[i]}].push_back(i);
      ve[tmp[i]].push_back(arr[i]);
    }
  }
  int uk = 0;
  for (int i = (0); i < (n); i++) {
    tour(tmp[i], i);
    out[sol].pop_back();
    if (out[sol].size()) {
      reverse(out[sol].begin(), out[sol].end());
      uk += out[sol].size();
      sol++;
    }
  }
  if (uk > s) {
    cout << "-1\n";
    return 0;
  }
  int m = min(sol, s - uk);
  for (int i = (0); i < (m); i++) {
    sol--;
    for (int x : out[sol]) t1.push_back(x);
    t2.push_back(out[sol][0]);
  }
  if (m == 1) sol++;
  if (m <= 1) {
    cout << sol << endl;
    for (int i = (0); i < (sol); i++) {
      cout << out[i].size() << "\n";
      for (int x : out[i]) cout << x + 1 << " ";
      cout << "\n";
    }
  } else {
    reverse(t2.begin(), t2.end());
    cout << sol + 2 << endl;
    cout << t1.size() << endl;
    for (int x : t1) cout << x + 1 << " ";
    cout << endl;
    cout << t2.size() << endl;
    for (int x : t2) cout << x + 1 << " ";
    cout << endl;
    for (int i = (0); i < (sol); i++) {
      cout << out[i].size() << "\n";
      for (int x : out[i]) cout << x + 1 << " ";
      cout << "\n";
    }
  }
  return 0;
}