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;
vector<int> new_val;
inline int Val(int first) {
  return lower_bound(new_val.begin(), new_val.end(), first) - new_val.begin();
}
void Move(vector<int> &ind, vector<int> &a) {
  int last = a[ind.back()];
  for (int i : ind) {
    swap(a[i], last);
  }
}
int n, s;
void Dfs(vector<vector<int> > &graph, int node, vector<bool> &visited) {
  visited[node] = true;
  for (int neighbor : graph[node]) {
    if (!visited[neighbor]) Dfs(graph, neighbor, visited);
  }
}
vector<int> Solve(int i, vector<vector<int> > &graph,
                  map<pair<int, int>, vector<int> > &v) {
  list<int> path = {i};
  for (auto it = path.begin(); it != path.end(); it++) {
    auto it2 = it;
    it2++;
    for (int node = *it; !graph[node].empty();) {
      int next = graph[node].back();
      graph[node].pop_back();
      node = next;
      path.insert(it2, node);
    }
  }
  vector<int> ans;
  for (auto it = path.begin(); it != path.end(); it++) {
    auto next = it;
    next++;
    if (next == path.end()) break;
    ans.push_back(v[{*it, *next}].back());
    v[{*it, *next}].pop_back();
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> s;
  vector<int> a(n + 1);
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    new_val.push_back(a[i]);
  }
  sort(new_val.begin(), new_val.end());
  vector<int> sorted = new_val;
  new_val.erase(unique(new_val.begin(), new_val.end()), new_val.end());
  for (int i = 0; i < n; i++) {
    sorted[i] = Val(sorted[i]);
  }
  for (int i = 1; i <= n; i++) {
    a[i] = Val(a[i]);
  }
  vector<vector<int> > graph(n + 1);
  int count = 0;
  map<pair<int, int>, vector<int> > v;
  vector<int> ind(n + 1);
  for (int i = 1; i <= n; i++) {
    if (sorted[i - 1] != a[i]) {
      count++;
      graph[sorted[i - 1]].push_back(a[i]);
      ind[a[i]] = i;
    }
  }
  if (count > s) {
    cout << -1;
    return 0;
  }
  vector<int> components;
  vector<bool> visited(n + 1);
  for (int i = 0; i <= n; i++) {
    if (graph[i].empty()) continue;
    if (!visited[i]) {
      Dfs(graph, i, visited);
      components.push_back(ind[i]);
    }
  }
  vector<vector<int> > moves = {};
  if (components.size() > 1 && s > count + 1) {
    moves.push_back({});
    for (int i = 0; i < min(s - count, int(components.size())); i++) {
      moves.back().push_back(components[i]);
    }
    Move(moves.back(), a);
  }
  for (int i = 0; i <= n; i++) {
    graph[i].clear();
    visited[i] = false;
  }
  for (int i = 1; i <= n; i++) {
    if (sorted[i - 1] != a[i]) {
      v[{sorted[i - 1], a[i]}].push_back(i);
      count++;
      graph[sorted[i - 1]].push_back(a[i]);
    }
  }
  for (int i = 0; i <= n; i++) {
    if (graph[i].empty()) continue;
    if (!visited[i]) {
      Dfs(graph, i, visited);
      moves.push_back(Solve(i, graph, v));
    }
  }
  cout << moves.size() << '\n';
  for (vector<int> &move : moves) {
    cout << move.size() << '\n';
    for (int i : move) {
      cout << i << ' ';
    }
    cout << '\n';
  }
}