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 M = 2e5 + 239;
int n, s, a[M], p[M];
pair<int, int> b[M];
int parent[M], r[M];
inline void init() {
  for (int i = 0; i < n; i++) {
    parent[i] = i;
    r[i] = 0;
  }
}
inline int find_set(int p) {
  if (parent[p] == p) return p;
  return (parent[p] = find_set(parent[p]));
}
inline void merge_set(int i, int j) {
  i = find_set(i);
  j = find_set(j);
  if (i == j) return;
  if (r[i] > r[j]) swap(i, j);
  if (r[i] == r[j]) r[j]++;
  parent[i] = j;
}
inline bool is_connect(int i, int j) { return (find_set(i) == find_set(j)); }
int t;
bool used[M];
vector<int> c[M];
void dfs(int x) {
  used[x] = true;
  c[t].push_back(x);
  if (!used[p[x]]) dfs(p[x]);
}
int main() {
  ios::sync_with_stdio(0);
  cin >> n >> s;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    b[i] = make_pair(a[i], i);
  }
  sort(b, b + n);
  for (int i = 0; i < n; i++) p[b[i].second] = i;
  for (int i = 0; i < n; i++)
    if (a[i] == b[i].first && p[i] != i) {
      p[b[i].second] = p[i];
      b[p[i]].second = b[i].second;
      p[i] = i;
      b[i].second = i;
    }
  init();
  for (int i = 0; i < n; i++) merge_set(p[i], i);
  int ls = -1;
  for (int i = 0; i < n; i++) {
    if (p[b[i].second] == b[i].second) continue;
    if (ls >= 0 && a[ls] == a[b[i].second]) {
      int x = ls;
      int y = b[i].second;
      if (is_connect(x, y)) continue;
      merge_set(x, y);
      swap(p[x], p[y]);
    }
    ls = b[i].second;
  }
  t = 0;
  for (int i = 0; i < n; i++)
    if (!used[i] && p[i] != i) {
      dfs(i);
      t++;
    }
  int kol = 0;
  for (int i = 0; i < t; i++) kol += (int)c[i].size();
  if (kol > s) {
    cout << "-1";
    return 0;
  }
  s -= kol;
  s = min(s, t);
  if (s <= 1) {
    cout << t << "\n";
    for (int i = 0; i < t; i++) {
      cout << c[i].size() << "\n";
      for (int x : c[i]) cout << (x + 1) << " ";
      cout << "\n";
    }
    return 0;
  }
  cout << (t - s + 2) << "\n";
  for (int i = 0; i < t - s; i++) {
    cout << c[i + s].size() << "\n";
    for (int x : c[i + s]) cout << (x + 1) << " ";
    cout << "\n";
    kol -= (int)c[i + s].size();
  }
  cout << kol << "\n";
  for (int i = 0; i < s; i++)
    for (int x : c[i]) cout << (x + 1) << " ";
  cout << "\n";
  cout << s << "\n";
  for (int i = s - 1; i >= 0; i--) cout << (c[i][0] + 1) << " ";
  cout << "\n";
  return 0;
}