#include <bits/stdc++.h>
using namespace std;
using ll = long long;
template <typename S, typename T>
void xmin(S &a, T const &b) {
  if (b < a) a = b;
}
template <typename S, typename T>
void xmax(S &a, T const &b) {
  if (b > a) a = b;
}
vector<vector<pair<int, int> > > g;
vector<int> eu;
vector<int> cured;
void rec(int u) {
  for (int &i = cured[u]; i < (int)g[u].size();) {
    int v = g[u][i].first;
    int x = g[u][i].second;
    ++i;
    rec(v);
    eu.push_back(x);
  }
}
signed main() {
  cin.tie(0);
  cout.tie(0);
  ios_base::sync_with_stdio(false);
  int n, s;
  cin >> n >> s;
  vector<int> a(n), b;
  for (auto &e : a) {
    cin >> e;
  }
  b = a;
  sort(b.begin(), b.end());
  for (auto &e : a) {
    e = lower_bound(b.begin(), b.end(), e) - b.begin();
  }
  b = a;
  sort(b.begin(), b.end());
  g.clear();
  g.resize(n);
  for (int i = 0; i < n; ++i) {
    if (a[i] != b[i]) {
      g[a[i]].emplace_back(b[i], i);
    }
  }
  vector<vector<int> > cyc;
  int tot = 0;
  cured.assign(n, 0);
  for (int i = 0; i < n; ++i) {
    eu.clear();
    rec(i);
    if (!eu.empty()) {
      cyc.push_back(eu);
      tot += eu.size();
    }
  }
  if (tot > s) {
    cout << "-1\n";
  } else {
    int spare = s - tot;
    if (spare > 2 && cyc.size() > 2) {
      xmin(spare, (int)cyc.size());
      vector<int> big, small;
      for (int i = 0; i < spare; ++i) {
        small.push_back(cyc.back().back());
        big.insert(big.end(), cyc.back().begin(), cyc.back().end());
        cyc.pop_back();
      }
      reverse(small.begin(), small.end());
      cyc.push_back(small);
      cyc.push_back(big);
    }
    cout << cyc.size() << "\n";
    for (auto const &e : cyc) {
      cout << e.size() << "\n";
      for (auto const &f : e) {
        cout << 1 + f << " ";
      }
      cout << "\n";
    }
  }
  return 0;
}

### Prompt

Your task is to create 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;
using ll = long long;
template <typename S, typename T>
void xmin(S &a, T const &b) {
  if (b < a) a = b;
}
template <typename S, typename T>
void xmax(S &a, T const &b) {
  if (b > a) a = b;
}
vector<vector<pair<int, int> > > g;
vector<int> eu;
vector<int> cured;
void rec(int u) {
  for (int &i = cured[u]; i < (int)g[u].size();) {
    int v = g[u][i].first;
    int x = g[u][i].second;
    ++i;
    rec(v);
    eu.push_back(x);
  }
}
signed main() {
  cin.tie(0);
  cout.tie(0);
  ios_base::sync_with_stdio(false);
  int n, s;
  cin >> n >> s;
  vector<int> a(n), b;
  for (auto &e : a) {
    cin >> e;
  }
  b = a;
  sort(b.begin(), b.end());
  for (auto &e : a) {
    e = lower_bound(b.begin(), b.end(), e) - b.begin();
  }
  b = a;
  sort(b.begin(), b.end());
  g.clear();
  g.resize(n);
  for (int i = 0; i < n; ++i) {
    if (a[i] != b[i]) {
      g[a[i]].emplace_back(b[i], i);
    }
  }
  vector<vector<int> > cyc;
  int tot = 0;
  cured.assign(n, 0);
  for (int i = 0; i < n; ++i) {
    eu.clear();
    rec(i);
    if (!eu.empty()) {
      cyc.push_back(eu);
      tot += eu.size();
    }
  }
  if (tot > s) {
    cout << "-1\n";
  } else {
    int spare = s - tot;
    if (spare > 2 && cyc.size() > 2) {
      xmin(spare, (int)cyc.size());
      vector<int> big, small;
      for (int i = 0; i < spare; ++i) {
        small.push_back(cyc.back().back());
        big.insert(big.end(), cyc.back().begin(), cyc.back().end());
        cyc.pop_back();
      }
      reverse(small.begin(), small.end());
      cyc.push_back(small);
      cyc.push_back(big);
    }
    cout << cyc.size() << "\n";
    for (auto const &e : cyc) {
      cout << e.size() << "\n";
      for (auto const &f : e) {
        cout << 1 + f << " ";
      }
      cout << "\n";
    }
  }
  return 0;
}

```