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;
struct ln {
  int val;
  ln* next;
};
vector<pair<int, int> >* nl;
vector<pair<int, int> > ar;
int* roi;
bool* btdt;
ln* fe(ln* ath, int cp) {
  ln* sv = NULL;
  ln* ov = NULL;
  for (int i = roi[cp]; i < nl[cp].size(); i = roi[cp]) {
    int np = nl[cp][i].first;
    int ind = nl[cp][i].second;
    if (btdt[ind]) {
      continue;
    }
    roi[cp]++;
    btdt[ind] = 1;
    ln* cn = new ln();
    cn->val = ind;
    cn->next = NULL;
    ln* vas = fe(cn, np);
    if (sv == NULL) {
      sv = cn;
      ov = vas;
      continue;
    }
    ath->next = cn;
    ath = vas;
  }
  if (sv != NULL) {
    ath->next = sv;
    ath = ov;
  }
  return ath;
}
int main() {
  cin.sync_with_stdio(0);
  cout.sync_with_stdio(0);
  int n, s;
  cin >> n >> s;
  int S = s;
  nl = new vector<pair<int, int> >[n];
  btdt = new bool[n];
  roi = new int[n];
  for (int i = 0; i < n; i++) {
    roi[i] = 0;
    vector<pair<int, int> > vpii;
    nl[i] = vpii;
    btdt[i] = 0;
    int a;
    cin >> a;
    ar.push_back(make_pair(a, i));
  }
  vector<pair<int, int> > br = ar;
  sort(br.begin(), br.end());
  unordered_map<int, int> nct;
  int cn = -1;
  int an = -1;
  for (int i = 0; i < n; i++) {
    if (br[i].first != an) {
      cn++;
      an = br[i].first;
      nct[an] = cn;
    }
    br[i].first = cn;
  }
  for (int i = 0; i < n; i++) {
    ar[i].first = nct[ar[i].first];
  }
  for (int i = 0; i < n; i++) {
    if (br[i].first == ar[i].first) {
      continue;
    }
    s--;
    nl[br[i].first].push_back(make_pair(ar[i].first, ar[i].second));
  }
  if (s < 0) {
    cout << -1 << "\n";
    return 0;
  }
  vector<ln*> atc;
  for (int i = 0; i < n; i++) {
    ln* tn = new ln();
    tn->val = -1;
    tn->next = NULL;
    ln* on = fe(tn, i);
    if (on != tn) {
      atc.push_back(tn->next);
    }
  }
  int noc = atc.size();
  int hmg = min(s, noc);
  vector<vector<int> > vas;
  if (hmg == 1) {
    hmg--;
  }
  if (hmg > 0) {
    if (S == 198000) {
    }
    vector<int> fv;
    vector<int> sv;
    for (int i = hmg - 1; i >= 0; i--) {
      fv.push_back(atc[i]->val);
    }
    for (int i = 0; i < hmg; i++) {
      int oi = i - 1;
      if (oi < 0) {
        oi += hmg;
      }
      sv.push_back(atc[oi]->val);
      ln* cn = atc[i]->next;
      while (cn != NULL) {
        sv.push_back(cn->val);
        cn = cn->next;
      }
    }
    vas.push_back(fv);
    vas.push_back(sv);
  }
  for (int i = hmg; i < atc.size(); i++) {
    vector<int> sv;
    ln* cn = atc[i];
    while (cn != NULL) {
      sv.push_back(cn->val);
      cn = cn->next;
    }
    vas.push_back(sv);
  }
  cout << vas.size() << "\n";
  for (int i = 0; i < vas.size(); i++) {
    cout << vas[i].size() << "\n";
    for (int j = 0; j < vas[i].size(); j++) {
      if (j > 0) {
        cout << " ";
      }
      cout << (vas[i][j] + 1);
    }
    cout << "\n";
  }
  cin >> n;
}