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, s, a[200079], p[200079];
vector<pair<int, int> > b(200079);
int rodic[200079], r[200079];
int najdisef(int v) {
  if (rodic[v] == v) return v;
  rodic[v] = najdisef(rodic[v]);
  return rodic[v];
}
void spojset(int i, int j) {
  i = najdisef(i);
  j = najdisef(j);
  if (i == j) return;
  if (r[i] > r[j]) swap(i, j);
  if (r[i] == r[j]) r[j]++;
  rodic[i] = j;
}
int t;
vector<int> u(200079, false);
vector<vector<int> > c(200079);
void dfs(int vr) {
  u[vr] = true;
  c[t].push_back(vr);
  if (!u[p[vr]]) dfs(p[vr]);
}
bool cmp(pair<int, int> p1, pair<int, int> p2) {
  if (p1.first == p2.first) return p1.second > p2.second;
  return p1.first > p2.first;
}
int main() {
  cin >> n >> s;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    pair<int, int> tmp;
    tmp.first = a[i];
    tmp.second = i;
    b[i] = tmp;
  }
  sort(b.begin(), b.begin() + 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;
    }
  }
  for (int i = 0; i < n; i++) {
    rodic[i] = i;
    r[i] = 0;
  }
  for (int i = 0; i < n; i++) spojset(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 va = ls, vb = b[i].second;
      if (najdisef(va) == najdisef(vb)) continue;
      spojset(va, vb);
      swap(p[va], p[vb]);
    }
    ls = b[i].second;
  }
  t = 0;
  for (int i = 0; i < n; i++)
    if (u[i] == 0 && p[i] != i) {
      dfs(i);
      t++;
    }
  int ans = 0;
  for (int i = 0; i < t; i++) ans += c[i].size();
  if (ans > s) {
    cout << "-1\n";
    return 0;
  }
  s -= ans;
  s = min(s, t);
  if (s <= 1) {
    cout << t << endl;
    for (int i = 0; i < t; i++) {
      cout << c[i].size() << endl;
      for (int j = 0; j < c[i].size(); j++) cout << c[i][j] + 1 << " ";
      cout << endl;
    }
    return 0;
  }
  cout << (t - s + 2) << endl;
  for (int i = 0; i < t - s; i++) {
    cout << c[i + s].size() << endl;
    for (int j = 0; j < c[i + s].size(); j++) cout << c[i + s][j] + 1 << " ";
    cout << endl;
    ans -= c[i + s].size();
  }
  cout << ans << endl;
  for (int i = 0; i < s; i++)
    for (int j = 0; j < c[i].size(); j++) cout << c[i][j] + 1 << " ";
  cout << endl;
  cout << s << endl;
  for (int i = s - 1; i >= 0; i--) cout << c[i][0] + 1 << " ";
  cout << "\n";
  return 0;
}