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 = 202020;
int N, S, val[MAXN], sorted[MAXN], begin_idx[MAXN], incl[MAXN], c = 0;
map<int, int> compress;
vector<int> adj[MAXN], cycle;
map<pair<int, int>, vector<int>> locs;
void recur(int n) {
  ;
  while (adj[n].size() > 0) {
    int nex = adj[n].back();
    adj[n].pop_back();
    recur(nex);
  };
  cycle.push_back(n);
}
vector<int> reconstruct(vector<int>& cycle) {
  ;
  vector<int> result;
  for (int i = 0; i < cycle.size(); ++i) {
    int nex = (i + 1) % cycle.size();
    pair<int, int> key = {cycle[i], cycle[nex]};
    vector<int>& opts = locs[key];
    ;
    assert(opts.size() > 0);
    result.push_back(opts.back());
    opts.pop_back();
  }
  return result;
}
int main() {
  scanf("%d %d", &N, &S);
  for (int i = 1; i <= N; ++i) {
    scanf("%d", &val[i]);
  }
  for (int i = 1; i <= N; ++i) {
    sorted[i] = val[i];
  }
  sort(sorted + 1, sorted + N + 1);
  for (int i = 1; i <= N; ++i) {
    if (compress.find(sorted[i]) == compress.end()) {
      compress[sorted[i]] = c++;
    }
  }
  for (int i = 1; i <= N; ++i) {
    val[i] = compress[val[i]];
    sorted[i] = val[i];
  }
  int nswaps = 0;
  sort(sorted + 1, sorted + N + 1);
  for (int i = 1; i <= N; ++i) {
    if (val[i] != sorted[i]) {
      adj[sorted[i]].push_back(val[i]);
      nswaps++;
    } else {
      incl[i] = 1;
    }
  }
  if (nswaps > S) {
    printf("-1\n");
    return 0;
  }
  for (int i = 1; i <= N; ++i) {
    if (begin_idx[sorted[i]] == 0) {
      begin_idx[sorted[i]] = i;
    }
  }
  for (int i = 1; i <= N; ++i) {
    ;
    if (locs.find({sorted[i], val[i]}) == locs.end()) {
      locs[{sorted[i], val[i]}] = vector<int>(1, i);
    } else {
      locs[{sorted[i], val[i]}].push_back(i);
    }
  };
  vector<vector<int>> cycles;
  for (int i = 0; i < c; ++i) {
    cycle.clear();
    ;
    recur(i);
    reverse(cycle.begin(), cycle.end());
    cycle.pop_back();
    if (cycle.size() > 0) {
      cycle = reconstruct(cycle);
      cycles.push_back(cycle);
    }
  };
  int to_merge = min(S - nswaps, (int)cycles.size());
  if (to_merge > 1) {
    vector<int> cycle1, cycle2;
    for (int i = 0; i < to_merge; ++i) {
      for (int j = 0; j < cycles.back().size(); ++j) {
        cycle1.push_back(cycles.back()[j]);
      }
      cycle2.push_back(cycles.back()[0]);
      cycles.pop_back();
    }
    reverse(cycle2.begin(), cycle2.end());
    cycles.push_back(cycle1);
    cycles.push_back(cycle2);
  };
  printf("%d\n", (int)cycles.size());
  for (int i = 0; i < cycles.size(); ++i) {
    printf("%d\n", (int)cycles[i].size());
    for (int j = 0; j < cycles[i].size(); ++j) {
      printf("%d", cycles[i][j]);
      if (j == cycles[i].size() - 1)
        printf("\n");
      else
        printf(" ");
    }
  }
  return 0;
}