#include <bits/stdc++.h>
using namespace std;
const long long inf = 1e18;
vector<bool> vis;
vector<pair<long long, long long>> eulerWalk(
    vector<vector<pair<long long, long long>>>& gr, vector<long long>& eu,
    int src, vector<long long>& D, vector<long long>& its) {
  vector<pair<long long, long long>> ret, s = {{src, -1}};
  D[src]++;
  while (!s.empty()) {
    auto [x, id] = s.back();
    int y, e;
    long long& it = its[x];
    int end = (int)gr[x].size();
    vis[x] = 1;
    if (it == end) {
      ret.emplace_back(x, id);
      s.pop_back();
      continue;
    }
    tie(y, e) = gr[x][it++];
    if (!eu[e]) {
      D[x]--, D[y]++;
      eu[e] = 1;
      s.emplace_back(y, e);
    }
  }
  return {ret.rbegin(), ret.rend()};
}
int main() {
  cin.tie(0);
  cin.sync_with_stdio(0);
  (cout << fixed).precision(15);
  int n;
  int s;
  cin >> n >> s;
  vector<long long> v(n);
  for (auto& i : (v)) cin >> i;
  vector<long long> o = v;
  sort(begin(o), end(o));
  map<int, int> conv;
  int k = 0;
  auto get_E = [&]() {
    conv.clear();
    vector<vector<pair<long long, long long>>> E;
    auto get = [&](int u) {
      if (conv.count(u) == 0) {
        conv[u] = conv.size();
        E.emplace_back();
      }
      return conv[u];
    };
    for (int i = (0); i < int(n); ++i)
      if (o[i] != v[i]) {
        ++k;
        int a = get(o[i]);
        int b = get(v[i]);
        E[a].emplace_back(b, i);
      }
    return E;
  };
  vector<vector<long long>> ans = {{}};
  auto E = get_E();
  set<int> seen;
  for (int i = (0); i < int(n); ++i)
    if (v[i] != o[i] && !seen.count(conv[v[i]])) {
      ans[0].push_back(i);
      queue<int> q;
      q.push(conv[v[i]]);
      seen.insert(conv[v[i]]);
      while (q.size()) {
        int u = q.front();
        q.pop();
        for (auto& w : (E[u]))
          if (!seen.count(w.first)) {
            q.push(w.first);
            seen.insert(w.first);
          }
      }
    }
  while (ans[0].size() && (int)ans[0].size() + k > s) ans[0].pop_back();
  if (ans[0].size() > 2) {
    for (int i = (1); i < int(ans[0].size()); ++i)
      swap(v[ans[0][i - 1]], v[ans[0][i]]);
    reverse(begin(ans[0]), end(ans[0]));
    s -= ans[0].size();
  } else {
    ans = {};
  }
  E = get_E();
  vis.assign(E.size(), 0);
  vector<long long> used(n);
  int m = E.size();
  vector<long long> D(m), its(m);
  for (int i = (0); i < int(E.size()); ++i)
    if (!vis[i]) {
      auto t = eulerWalk(E, used, i, D, its);
      t.erase(t.begin());
      s -= t.size();
      ans.emplace_back();
      for (auto& i : (t)) ans.back().push_back(i.second);
    }
  if (s < 0) {
    cout << -1 << endl;
  } else {
    cout << ans.size() << endl;
    for (auto& i : (ans)) {
      cout << i.size() << endl;
      for (auto& j : (i)) cout << j + 1 << " ";
      cout << endl;
    }
  }
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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 long long inf = 1e18;
vector<bool> vis;
vector<pair<long long, long long>> eulerWalk(
    vector<vector<pair<long long, long long>>>& gr, vector<long long>& eu,
    int src, vector<long long>& D, vector<long long>& its) {
  vector<pair<long long, long long>> ret, s = {{src, -1}};
  D[src]++;
  while (!s.empty()) {
    auto [x, id] = s.back();
    int y, e;
    long long& it = its[x];
    int end = (int)gr[x].size();
    vis[x] = 1;
    if (it == end) {
      ret.emplace_back(x, id);
      s.pop_back();
      continue;
    }
    tie(y, e) = gr[x][it++];
    if (!eu[e]) {
      D[x]--, D[y]++;
      eu[e] = 1;
      s.emplace_back(y, e);
    }
  }
  return {ret.rbegin(), ret.rend()};
}
int main() {
  cin.tie(0);
  cin.sync_with_stdio(0);
  (cout << fixed).precision(15);
  int n;
  int s;
  cin >> n >> s;
  vector<long long> v(n);
  for (auto& i : (v)) cin >> i;
  vector<long long> o = v;
  sort(begin(o), end(o));
  map<int, int> conv;
  int k = 0;
  auto get_E = [&]() {
    conv.clear();
    vector<vector<pair<long long, long long>>> E;
    auto get = [&](int u) {
      if (conv.count(u) == 0) {
        conv[u] = conv.size();
        E.emplace_back();
      }
      return conv[u];
    };
    for (int i = (0); i < int(n); ++i)
      if (o[i] != v[i]) {
        ++k;
        int a = get(o[i]);
        int b = get(v[i]);
        E[a].emplace_back(b, i);
      }
    return E;
  };
  vector<vector<long long>> ans = {{}};
  auto E = get_E();
  set<int> seen;
  for (int i = (0); i < int(n); ++i)
    if (v[i] != o[i] && !seen.count(conv[v[i]])) {
      ans[0].push_back(i);
      queue<int> q;
      q.push(conv[v[i]]);
      seen.insert(conv[v[i]]);
      while (q.size()) {
        int u = q.front();
        q.pop();
        for (auto& w : (E[u]))
          if (!seen.count(w.first)) {
            q.push(w.first);
            seen.insert(w.first);
          }
      }
    }
  while (ans[0].size() && (int)ans[0].size() + k > s) ans[0].pop_back();
  if (ans[0].size() > 2) {
    for (int i = (1); i < int(ans[0].size()); ++i)
      swap(v[ans[0][i - 1]], v[ans[0][i]]);
    reverse(begin(ans[0]), end(ans[0]));
    s -= ans[0].size();
  } else {
    ans = {};
  }
  E = get_E();
  vis.assign(E.size(), 0);
  vector<long long> used(n);
  int m = E.size();
  vector<long long> D(m), its(m);
  for (int i = (0); i < int(E.size()); ++i)
    if (!vis[i]) {
      auto t = eulerWalk(E, used, i, D, its);
      t.erase(t.begin());
      s -= t.size();
      ans.emplace_back();
      for (auto& i : (t)) ans.back().push_back(i.second);
    }
  if (s < 0) {
    cout << -1 << endl;
  } else {
    cout << ans.size() << endl;
    for (auto& i : (ans)) {
      cout << i.size() << endl;
      for (auto& j : (i)) cout << j + 1 << " ";
      cout << endl;
    }
  }
}

```