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 chrono;
const int infinity = (int)1e9 + 42;
const int64_t llInfinity = (int64_t)1e18 + 256;
const int module = (int)1e9 + 7;
const long double eps = 1e-8;
mt19937_64 randGen(system_clock().now().time_since_epoch().count());
inline void raiseError(string errorCode) {
  cerr << "Error : " << errorCode << endl;
  exit(42);
}
signed main() {
  ios_base::sync_with_stdio(false);
  int n, s;
  cin >> n >> s;
  vector<int> v(n);
  map<int, int> kompr;
  for (int i = 0; i < n; i++) {
    cin >> v[i];
    kompr[v[i]] = 42;
  }
  int k = 0;
  for (auto &it : kompr) {
    it.second = k++;
  }
  for (int i = 0; i < n; i++) {
    v[i] = kompr[v[i]];
  }
  auto w = v;
  sort(w.begin(), w.end());
  vector<vector<int> > g(k);
  map<pair<int, int>, vector<int> > indices;
  for (int i = 0; i < n; i++) {
    if (w[i] == v[i]) {
      continue;
    }
    s--;
    g[w[i]].push_back(v[i]);
    indices[{w[i], v[i]}].push_back(i);
  }
  if (s < 0) {
    cout << -1 << "\n";
    return 0;
  }
  vector<int> pass;
  function<void(int)> dfs = [&](int v) {
    while (!g[v].empty()) {
      int to = g[v].back();
      g[v].pop_back();
      dfs(to);
    }
    pass.push_back(v);
  };
  vector<vector<int> > ans;
  for (int i = 0; i < k; i++) {
    if (g[i].empty()) {
      continue;
    }
    pass.clear();
    dfs(i);
    ans.emplace_back();
    for (int j = 1; j < (int)pass.size(); j++) {
      auto &vec = indices[{pass[j], pass[j - 1]}];
      assert(!vec.empty());
      ans.back().push_back(vec.back());
      vec.pop_back();
    }
    reverse(ans.back().begin(), ans.back().end());
  }
  int canUnite = min(s, (int)ans.size());
  if (canUnite > 2) {
    vector<int> united;
    vector<int> loop2;
    for (int i = 0; i < canUnite; i++) {
      united.insert(united.end(), ans.back().begin(), ans.back().end());
      loop2.push_back(ans.back()[0]);
      ans.pop_back();
    }
    ans.push_back(united);
    reverse(loop2.begin(), loop2.end());
    ans.push_back(loop2);
  }
  cout << ans.size() << "\n";
  for (auto vec : ans) {
    cout << vec.size() << "\n";
    for (auto it : vec) {
      cout << it + 1 << " ";
    }
    cout << "\n";
  }
  return 0;
}