#include <bits/stdc++.h>
using namespace std;
int a[200010], b[200010], nxt[200010];
bool vis[200010];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n, s;
  cin >> n >> s;
  for (int i = 1; i <= n; i++) cin >> a[i], b[i] = a[i];
  sort(b + 1, b + n + 1);
  vector<int> v;
  for (int i = 1; i <= n; i++)
    if (a[i] != b[i]) v.push_back(i);
  if (v.size() > s) {
    puts("-1");
    return 0;
  }
  vector<int> u = v;
  sort(v.begin(), v.end(), [&](int i, int j) { return a[i] < a[j]; });
  for (int i = 0; i < v.size(); i++) nxt[v[i]] = u[i];
  vector<int> f(n + 1);
  for (int i = 1; i <= n; i++) f[i] = i;
  function<int(int)> find = [&](int x) {
    return x == f[x] ? x : f[x] = find(f[x]);
  };
  for (int i = 0; i < v.size(); i++) {
    int x = find(u[i]), y = find(v[i]);
    f[x] = y;
  }
  for (int i = 0, prv = -1; i < v.size(); i++) {
    if (a[v[i]] == prv) {
      int x = find(v[i]), y = find(v[i - 1]);
      if (x != y) {
        swap(nxt[v[i]], nxt[v[i - 1]]);
        f[x] = y;
      }
    }
    prv = a[v[i]];
  }
  vector<vector<int> > R;
  for (int i = 0; i < v.size(); i++) {
    int x = v[i];
    if (vis[x]) continue;
    vector<int> r;
    while (vis[x] == 0) r.push_back(x), vis[x] = 1, x = nxt[x];
    R.push_back(r);
  }
  int k = min(s - u.size(), R.size());
  if (k <= 2) k = 0;
  int ans = R.size() - k;
  if (k > 2) ans += 2;
  cout << ans << endl;
  if (k > 2) {
    cout << 2 * k << endl;
    for (int i = 0; i < k; i++) cout << R[i][0] << ' ' << R[i][1] << ' ';
    cout << endl;
    int sum = 0;
    for (int i = 0; i < k; i++) sum += R[i].size() - 1;
    cout << sum << endl;
    for (int i = k - 1; i >= 0; i--) {
      for (int j = 2; j < R[i].size(); j++) cout << R[i][j] << ' ';
      cout << R[i][0] << ' ';
    }
    cout << endl;
  }
  for (int i = k; i < R.size(); i++) {
    cout << R[i].size() << endl;
    for (int j = 0; j < R[i].size(); j++) cout << R[i][j] << ' ';
    cout << endl;
  }
}

### Prompt

Generate a cpp solution to the following 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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int a[200010], b[200010], nxt[200010];
bool vis[200010];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n, s;
  cin >> n >> s;
  for (int i = 1; i <= n; i++) cin >> a[i], b[i] = a[i];
  sort(b + 1, b + n + 1);
  vector<int> v;
  for (int i = 1; i <= n; i++)
    if (a[i] != b[i]) v.push_back(i);
  if (v.size() > s) {
    puts("-1");
    return 0;
  }
  vector<int> u = v;
  sort(v.begin(), v.end(), [&](int i, int j) { return a[i] < a[j]; });
  for (int i = 0; i < v.size(); i++) nxt[v[i]] = u[i];
  vector<int> f(n + 1);
  for (int i = 1; i <= n; i++) f[i] = i;
  function<int(int)> find = [&](int x) {
    return x == f[x] ? x : f[x] = find(f[x]);
  };
  for (int i = 0; i < v.size(); i++) {
    int x = find(u[i]), y = find(v[i]);
    f[x] = y;
  }
  for (int i = 0, prv = -1; i < v.size(); i++) {
    if (a[v[i]] == prv) {
      int x = find(v[i]), y = find(v[i - 1]);
      if (x != y) {
        swap(nxt[v[i]], nxt[v[i - 1]]);
        f[x] = y;
      }
    }
    prv = a[v[i]];
  }
  vector<vector<int> > R;
  for (int i = 0; i < v.size(); i++) {
    int x = v[i];
    if (vis[x]) continue;
    vector<int> r;
    while (vis[x] == 0) r.push_back(x), vis[x] = 1, x = nxt[x];
    R.push_back(r);
  }
  int k = min(s - u.size(), R.size());
  if (k <= 2) k = 0;
  int ans = R.size() - k;
  if (k > 2) ans += 2;
  cout << ans << endl;
  if (k > 2) {
    cout << 2 * k << endl;
    for (int i = 0; i < k; i++) cout << R[i][0] << ' ' << R[i][1] << ' ';
    cout << endl;
    int sum = 0;
    for (int i = 0; i < k; i++) sum += R[i].size() - 1;
    cout << sum << endl;
    for (int i = k - 1; i >= 0; i--) {
      for (int j = 2; j < R[i].size(); j++) cout << R[i][j] << ' ';
      cout << R[i][0] << ' ';
    }
    cout << endl;
  }
  for (int i = k; i < R.size(); i++) {
    cout << R[i].size() << endl;
    for (int j = 0; j < R[i].size(); j++) cout << R[i][j] << ' ';
    cout << endl;
  }
}

```