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 namespace rel_ops;
using ll = int64_t;
using Pii = pair<int, int>;
using ull = uint64_t;
using Vi = vector<int>;
void run();
int main() {
  cin.sync_with_stdio(0);
  cin.tie(0);
  cout << fixed << setprecision(10);
  run();
  return 0;
}
vector<vector<Pii>> G;
vector<Vi> cycles;
void dfs(int v) {
  while (!G[v].empty()) {
    Pii e = G[v].back();
    G[v].pop_back();
    dfs(e.first);
    cycles.back().push_back(e.second);
  }
}
void run() {
  int n, upper;
  cin >> n >> upper;
  Vi elems(n);
  for (auto& e : (elems)) cin >> e;
  Vi sorted(n);
  iota((sorted).begin(), (sorted).end(), 0);
  sort((sorted).begin(), (sorted).end(),
       [&](int l, int r) { return elems[l] < elems[r]; });
  int last = elems[sorted[0]], k = 0;
  for (int i = (0); i < (n); i++) {
    int e = sorted[i];
    if (elems[e] != last) k++;
    last = elems[e];
    elems[e] = k;
  }
  k++;
  G.resize(k);
  for (int i = (0); i < (n); i++)
    if (elems[i] != elems[sorted[i]]) {
      G[elems[i]].push_back({elems[sorted[i]], i});
    }
  int len = 0;
  for (int i = (0); i < (k); i++)
    if (!G[i].empty()) {
      cycles.emplace_back();
      dfs(i);
      len += int((cycles.back()).size());
    }
  if (len > upper) {
    cout << "-1\n";
    return;
  }
  int toMerge = min(int((cycles).size()), upper - len);
  if (toMerge > 2) {
    Vi one, two;
    for (int i = (0); i < (toMerge); i++) {
      one.insert(one.end(), (cycles.back()).begin(), (cycles.back()).end());
      two.push_back(cycles.back()[0]);
      cycles.pop_back();
    }
    reverse((two).begin(), (two).end());
    cycles.push_back(move(one));
    cycles.push_back(move(two));
  }
  cout << int((cycles).size()) << '\n';
  for (auto& c : (cycles)) {
    cout << int((c).size()) << '\n';
    for (auto& first : (c)) cout << first + 1 << ' ';
    cout << '\n';
  }
}