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;
int a[200010];
int b[200010];
vector<pair<int, int> > graf[200010];
vector<int> tura;
vector<int> sz;
vector<vector<int> > sol;
vector<vector<int> > rj;
vector<int> r;
int bio[200010];
int siz = 0;
void dfs(int x) {
  bio[x] = 1;
  while (!graf[x].empty()) {
    int sus = graf[x][(int)graf[x].size() - 1].first;
    int br = graf[x][(int)graf[x].size() - 1].second;
    graf[x].pop_back();
    dfs(sus);
    tura.push_back(br);
  }
}
void sazimanje() {
  for (int i = 1; i <= n; i++) sz.push_back(a[i]);
  sort(sz.begin(), sz.end());
  sz.erase(unique(sz.begin(), sz.end()), sz.end());
  siz = (int)sz.size() - 1;
  for (int i = 1; i <= n; i++) {
    a[i] = lower_bound(sz.begin(), sz.end(), a[i]) - sz.begin();
    b[i - 1] = a[i];
  }
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
  }
  sazimanje();
  sort(b, b + n);
  int cnt = 0;
  for (int i = n - 1; i >= 0; i--) {
    b[i + 1] = b[i];
    b[i] = 0;
    if (a[i + 1] != b[i + 1]) {
      cnt++;
      graf[b[i + 1]].push_back(make_pair(a[i + 1], i + 1));
    }
  }
  if (cnt > k) {
    printf("-1");
    return 0;
  }
  for (int i = 1; i <= siz; i++) {
    if (bio[i] == 0) {
      tura.clear();
      dfs(i);
      reverse(tura.begin(), tura.end());
      if (tura.size() >= 1) sol.push_back(tura);
    }
  }
  int sad = min((int)sol.size(), k - cnt);
  if (sad > 1) {
    for (int i = sad - 1; i >= 0; i--) {
      r.push_back(sol[i].back());
    }
    rj.push_back(r);
    r.clear();
    for (int i = 0; i < sad; i++) {
      for (int j = 0; j < sol[i].size(); j++) {
        r.push_back(sol[i][j]);
      }
    }
    rj.push_back(r);
    for (int i = sad; i < sol.size(); i++) {
      rj.push_back(sol[i]);
    }
    swap(sol, rj);
  }
  printf("%d\n", sol.size());
  for (int i = 0; i < sol.size(); i++) {
    printf("%d\n", sol[i].size());
    for (int j = 0; j < sol[i].size(); j++) printf("%d ", sol[i][j]);
    printf("\n");
  }
  return 0;
}