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 = 300000;
map<int, int> mapa;
map<pair<int, int>, vector<int>> pos;
vector<int> g[MAXN];
int ptr[MAXN];
int used[MAXN];
void euler(int v, vector<int> &res) {
  used[v] = true;
  for (; ptr[v] < (int)(g[v]).size();) {
    ++ptr[v];
    int u = g[v][ptr[v] - 1];
    euler(u, res);
    res.push_back(u);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n, s;
  cin >> n >> s;
  int k = 0;
  vector<int> a(n);
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
  }
  vector<int> b = a;
  sort((b).begin(), (b).end());
  int m = 0;
  for (int i = 0; i < n; ++i) {
    if (a[i] == b[i]) {
      continue;
    }
    ++m;
    if (!mapa.count(b[i])) {
      mapa[b[i]] = k++;
    }
  }
  if (m > s) {
    cout << -1 << endl;
    return 0;
  }
  for (int i = 0; i < n; ++i) {
    if (a[i] == b[i]) {
      continue;
    }
    a[i] = mapa[a[i]];
    b[i] = mapa[b[i]];
    g[b[i]].push_back(a[i]);
    pos[{b[i], a[i]}].push_back(i);
  }
  vector<vector<int>> cycles;
  for (int i = 0; i < k; ++i) {
    if (!used[i]) {
      vector<int> arr;
      euler(i, arr);
      reverse((arr).begin(), (arr).end());
      cycles.push_back({});
      for (int i = 0; i < (int)(arr).size(); ++i) {
        int j = (i + 1) % (int)(arr).size();
        cycles.back().push_back(pos[{arr[i], arr[j]}].back());
        pos[{arr[i], arr[j]}].pop_back();
      }
    }
  }
  vector<vector<int>> res;
  if (s - m > 1 && (int)(cycles).size() > 1) {
    int len = min((int)(cycles).size(), s - m);
    res.push_back({});
    vector<int> newcycle;
    for (int i = (int)(cycles).size() - len; i < (int)(cycles).size(); ++i) {
      res.back().push_back(cycles[i].back());
      for (int j : cycles[i]) {
        newcycle.push_back(j);
      }
    }
    reverse((res.back()).begin(), (res.back()).end());
    for (int i = 0; i < len; ++i) {
      cycles.pop_back();
    }
    cycles.push_back(newcycle);
  }
  for (int i = 0; i < (int)(cycles).size(); ++i) {
    res.push_back(cycles[i]);
  }
  cout << (int)(res).size() << endl;
  for (int i = 0; i < (int)(res).size(); ++i) {
    cout << (int)(res[i]).size() << endl;
    for (int j : res[i]) {
      cout << j + 1 << " ";
    }
    cout << endl;
  }
}