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;
using cat = long long;
void DFS(int R, vector<vector<int> >& G, vector<bool>& vis, vector<int>& cyc) {
  vis[R] = true;
  while (!G[R].empty()) {
    int v = G[R].back();
    G[R].pop_back();
    DFS(v, G, vis, cyc);
  }
  cyc.push_back(R);
}
int main() {
  cin.sync_with_stdio(0);
  cin.tie(0);
  cout << fixed << setprecision(10);
  int N, S;
  cin >> N >> S;
  vector<int> A(N);
  for (int i = 0; i < N; i++) cin >> A[i];
  vector<int> As(A);
  sort(begin(As), end(As));
  int wrong = 0;
  for (int i = 0; i < N; i++)
    if (A[i] != As[i]) wrong++;
  if (S < wrong) {
    cout << "-1\n";
    return 0;
  }
  map<int, int> M;
  for (int i = 0; i < N; i++) M[A[i]] = 0;
  int m = 0;
  for (auto it = M.begin(); it != M.end(); it++) it->second = m++;
  for (int i = 0; i < N; i++) A[i] = M[A[i]], As[i] = M[As[i]];
  vector<vector<int> > G(N + m);
  for (int i = 0; i < N; i++)
    if (A[i] != As[i]) G[i].push_back(A[i] + N);
  for (int i = 0; i < N; i++)
    if (A[i] != As[i]) G[As[i] + N].push_back(i);
  vector<bool> vis(N + m, false);
  vector<vector<int> > euc;
  for (int i = 0; i < N; i++)
    if (!vis[i] && A[i] != As[i]) {
      vector<int> c;
      DFS(i, G, vis, c);
      euc.push_back(vector<int>());
      reverse(begin(c), end(c));
      c.pop_back();
      for (auto it = c.begin(); it != c.end(); it++)
        if (*it < N) euc.back().push_back(*it);
    }
  int num_merged = min((int)euc.size(), S - wrong);
  if (num_merged <= 2) num_merged = 0;
  vector<vector<int> > ans;
  for (int i = num_merged; i < (int)euc.size(); i++) ans.push_back(euc[i]);
  if (num_merged > 0) {
    vector<int> mergedc;
    for (int i = 0; i < num_merged; i++)
      for (auto it = euc[i].begin(); it != euc[i].end(); it++)
        mergedc.push_back(*it);
    ans.push_back(mergedc);
    vector<int> failc;
    for (int i = 0; i < num_merged; i++) failc.push_back(euc[i][0]);
    reverse(begin(failc), end(failc));
    ans.push_back(failc);
  }
  cout << ans.size() << "\n";
  for (int i = 0; i < (int)ans.size(); i++) {
    cout << ans[i].size();
    for (auto it = ans[i].begin(); it != ans[i].end(); it++)
      cout << " " << *it + 1;
    cout << "\n";
  }
  return 0;
}