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;
int n, s;
int a[200100];
int sorted[200100];
int been[200100];
map<int, list<int>> alive;
list<list<int>> sol;
void dfs(int val) {
  if (alive[val].empty()) return;
  int k = alive[val].front();
  alive[val].pop_front();
  been[k] = true;
  dfs(a[k]);
  sol.back().push_front(k);
  if (!alive[val].empty()) dfs(val);
}
int main() {
  cin >> n >> s;
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < n; i++) sorted[i] = a[i];
  sort(sorted, sorted + n);
  for (int i = 0; i < n; i++) {
    if (a[i] == sorted[i])
      been[i] = true;
    else {
      s--;
      alive[sorted[i]].push_back(i);
    }
  }
  if (s < 0) {
    cout << -1 << endl;
    return 0;
  }
  for (int i = 0; i < n; i++)
    if (!been[i]) {
      sol.push_back(list<int>());
      dfs(sorted[i]);
    }
  if (s > 2 && sol.size() > 2) {
    list<int> new_cycle;
    list<int> rev_cycle;
    int num = min((size_t)s, sol.size());
    for (int w = 0; w < num; w++) {
      list<int> &l = sol.back();
      rev_cycle.push_front(l.front());
      for (auto x : sol.back()) new_cycle.push_back(x);
      sol.pop_back();
    }
    sol.push_back(new_cycle);
    sol.push_back(rev_cycle);
  }
  cout << sol.size() << endl;
  for (list<int> &l : sol) {
    cout << l.size() << endl;
    for (int x : l) cout << x + 1 << " ";
    cout << endl;
  }
  return 0;
}