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, K, a[200100];
map<int, int> mp1;
map<int, vector<int> > mp2;
vector<vector<int> > ans;
vector<int> vec;
void dfs(int x) {
  for (int i; !mp2[x].empty(); vec.push_back(i)) {
    i = mp2[x].back();
    mp2[x].pop_back();
    dfs(a[i]);
  }
  mp2.erase(x);
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> K;
  for (int i = 1; i <= n; ++i) cin >> a[i], mp1[a[i]]++;
  int sum = 1;
  for (auto i : mp1) {
    for (int j = sum; j < sum + i.second; ++j)
      if (a[j] != i.first) mp2[i.first].push_back(j);
    sum += i.second;
  }
  for (; !mp2.empty();) {
    vec.clear();
    dfs(mp2.begin()->first);
    reverse(vec.begin(), vec.end());
    ans.push_back(vec);
    K -= vec.size();
  }
  if (K < 0) {
    cout << -1 << '\n';
    return 0;
  }
  K = min(K, (int)ans.size());
  cout << (ans.size() - max(0, K - 2)) << '\n';
  if (K >= 3) {
    int len = 0;
    for (int i = 0; i < K; ++i) len += ans[i].size();
    cout << len << '\n';
    for (int i = 0; i < K; ++i)
      for (int j : ans[i]) cout << j << ' ';
    cout << '\n' << K << '\n';
    for (int i = K - 1; ~i; --i) cout << ans[i][0] << ' ';
    cout << '\n';
  } else
    K = 0;
  for (int i = K; i < (int)ans.size(); ++i) {
    cout << ans[i].size() << '\n';
    for (int j : ans[i]) cout << j << ' ';
    cout << '\n';
  }
  return 0;
}