#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 25;
int a[N], b[N], c[N], cycsz;
vector<int> g[N], cyc[N];
void dfs(int v) {
  while (!g[v].empty()) {
    int u = g[v].back();
    g[v].pop_back();
    dfs(a[u]);
    cyc[cycsz].push_back(u);
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout << setprecision(32);
  int n, s;
  cin >> n >> s;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    b[i] = a[i];
  }
  sort(b, b + n);
  memcpy(c, b, sizeof(int) * n);
  int sz = unique(c, c + n) - c;
  for (int i = 0; i < n; i++) {
    a[i] = lower_bound(c, c + sz, a[i]) - c;
    b[i] = lower_bound(c, c + sz, b[i]) - c;
    if (a[i] == b[i]) continue;
    g[b[i]].push_back(i);
  }
  cycsz = 0;
  for (int i = 0; i < sz; i++) {
    if (g[i].empty()) continue;
    dfs(i);
    cycsz++;
  }
  int sum = 0;
  for (int i = 0; i < cycsz; i++) {
    sum += cyc[i].size();
    reverse(cyc[i].begin(), cyc[i].end());
  }
  if (sum > s) {
    cout << -1 << '\n';
    exit(0);
  }
  int x = min(cycsz, s - sum);
  vector<vector<int> > ans;
  if (x) {
    vector<int> perm1, perm2;
    for (int i = 0; i < x; i++) {
      for (auto y : cyc[i]) {
        perm1.push_back(y);
      }
      perm2.push_back(cyc[i][0]);
    }
    reverse(perm2.begin(), perm2.end());
    ans.push_back(perm1);
    if (x > 1) ans.push_back(perm2);
  }
  for (int i = x; i < cycsz; i++) {
    ans.push_back(cyc[i]);
  }
  cout << ans.size() << '\n';
  for (auto vec : ans) {
    cout << vec.size() << '\n';
    for (auto y : vec) {
      cout << y + 1 << " ";
    }
    cout << '\n';
  }
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 25;
int a[N], b[N], c[N], cycsz;
vector<int> g[N], cyc[N];
void dfs(int v) {
  while (!g[v].empty()) {
    int u = g[v].back();
    g[v].pop_back();
    dfs(a[u]);
    cyc[cycsz].push_back(u);
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout << setprecision(32);
  int n, s;
  cin >> n >> s;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    b[i] = a[i];
  }
  sort(b, b + n);
  memcpy(c, b, sizeof(int) * n);
  int sz = unique(c, c + n) - c;
  for (int i = 0; i < n; i++) {
    a[i] = lower_bound(c, c + sz, a[i]) - c;
    b[i] = lower_bound(c, c + sz, b[i]) - c;
    if (a[i] == b[i]) continue;
    g[b[i]].push_back(i);
  }
  cycsz = 0;
  for (int i = 0; i < sz; i++) {
    if (g[i].empty()) continue;
    dfs(i);
    cycsz++;
  }
  int sum = 0;
  for (int i = 0; i < cycsz; i++) {
    sum += cyc[i].size();
    reverse(cyc[i].begin(), cyc[i].end());
  }
  if (sum > s) {
    cout << -1 << '\n';
    exit(0);
  }
  int x = min(cycsz, s - sum);
  vector<vector<int> > ans;
  if (x) {
    vector<int> perm1, perm2;
    for (int i = 0; i < x; i++) {
      for (auto y : cyc[i]) {
        perm1.push_back(y);
      }
      perm2.push_back(cyc[i][0]);
    }
    reverse(perm2.begin(), perm2.end());
    ans.push_back(perm1);
    if (x > 1) ans.push_back(perm2);
  }
  for (int i = x; i < cycsz; i++) {
    ans.push_back(cyc[i]);
  }
  cout << ans.size() << '\n';
  for (auto vec : ans) {
    cout << vec.size() << '\n';
    for (auto y : vec) {
      cout << y + 1 << " ";
    }
    cout << '\n';
  }
  return 0;
}

```