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 = 1 << 20;
int arr[MAXN], srt[MAXN];
bool bio[MAXN];
vector<pair<int, int>> edges[MAXN];
vector<vector<int>> cyc;
void euler(int i) {
  bio[i] = true;
  while (!edges[i].empty()) {
    auto x = edges[i].back();
    edges[i].pop_back();
    euler(x.first);
    cyc.back().push_back(x.second);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int n, s, m, i, j;
  unordered_map<int, int> cpr;
  cin >> n >> s;
  for (i = 0; i < n; ++i) {
    cin >> arr[i];
    srt[i] = arr[i];
  }
  sort(srt, srt + n);
  for (i = j = 0; i < n; ++i) {
    if (i && srt[i - 1] != srt[i]) {
      ++j;
      cpr[srt[i]] = j;
    }
  }
  m = j + 1;
  for (i = 0; i < n; ++i) {
    arr[i] = cpr[arr[i]];
    srt[i] = cpr[srt[i]];
  }
  for (i = 0; i < n; ++i)
    if (arr[i] != srt[i]) edges[arr[i]].push_back({srt[i], i});
  for (i = 0; i < m; ++i)
    if (!bio[i]) {
      cyc.emplace_back();
      euler(i);
      if (cyc.back().empty())
        cyc.pop_back();
      else
        s -= cyc.back().size();
    }
  if (s < 0) {
    cout << "-1\n";
    return 0;
  }
  if (s <= 2 || cyc.size() <= 2) {
    cout << cyc.size() << '\n';
    for (const auto& x : cyc) {
      cout << x.size() << '\n';
      for (int y : x) cout << y + 1 << ' ';
      cout << '\n';
    }
    return 0;
  }
  if (s > cyc.size()) s = cyc.size();
  cout << cyc.size() - s + 2 << '\n';
  m = 0;
  for (i = 0; i < s; ++i) {
    m += cyc[i].size();
  }
  cout << m << '\n';
  for (i = 0; i < s; ++i) {
    for (int y : cyc[i]) cout << y + 1 << ' ';
  }
  cout << s << '\n';
  for (i = s - 1; i >= 0; --i) {
    cout << cyc[i].front() + 1 << ' ';
  }
  cout << '\n';
  for (i = s; i < cyc.size(); ++i) {
    cout << cyc[i].size() << '\n';
    for (int y : cyc[i]) cout << y + 1 << ' ';
    cout << '\n';
  }
}