#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 5;
int n, s, a[N], aa[N], g[N], p[N], i, j;
int gfa(int x) { return g[x] == x ? x : g[x] = gfa(g[x]); }
bool bb[N], vi[N];
set<int> S;
unordered_map<int, int> be, en;
unordered_map<int, vector<int>> mp;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> s;
  for (i = 1; i <= n; ++i) cin >> a[i];
  memcpy(aa + 1, a + 1, n << 2);
  sort(aa + 1, aa + n + 1);
  for (i = 1; i <= n; ++i) {
    bb[i] = a[i] == aa[i], g[i] = i;
    if (!bb[i]) S.insert(i);
  }
  int ss = S.size();
  if (ss > s) {
    cout << -1 << endl;
    return 0;
  }
  for (i = 1; i <= n; ++i) en[aa[i]] = i;
  for (i = n; i; --i) be[aa[i]] = i;
  for (i = 1; i <= n; ++i)
    if (!bb[i])
      p[i] = *S.lower_bound(be[a[i]]), S.erase(p[i]), g[gfa(p[i])] = gfa(i);
  for (i = 1; i <= n; ++i)
    if (!bb[i])
      mp[a[i]].push_back(i);
    else
      p[i] = i;
  for (auto u : mp) {
    auto v = u.second;
    for (i = 1; i < v.size(); ++i)
      if (gfa(v[i]) != gfa(v[0]))
        g[gfa(v[i])] = gfa(v[0]), swap(p[v[i]], p[v[0]]);
  }
  vector<vector<int>> ans;
  for (i = 1; i <= n; ++i)
    if (p[i] != i && !vi[i]) {
      vector<int> ve;
      for (j = i; vi[j] = 1, ve.push_back(j), p[j] != i; j = p[j])
        ;
      ans.push_back(ve);
    }
  if (ss + 3 <= s && ans.size() > 1) {
    vector<int> v1, v2, v3;
    for (i = 1; ss + i <= s && !ans.empty(); ++i) {
      v3 = ans.back();
      ans.pop_back();
      v1.insert(v1.end(), v3.begin(), v3.end());
      v2.push_back(v3[0]);
    }
    reverse(v2.begin(), v2.end());
    ans.push_back(v1);
    ans.push_back(v2);
  }
  cout << ans.size() << '\n';
  for (auto u : ans) {
    cout << u.size() << '\n';
    for (int x : u) cout << x << ' ';
    cout << '\n';
  }
  return 0;
}

### 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;
const int N = 2e5 + 5;
int n, s, a[N], aa[N], g[N], p[N], i, j;
int gfa(int x) { return g[x] == x ? x : g[x] = gfa(g[x]); }
bool bb[N], vi[N];
set<int> S;
unordered_map<int, int> be, en;
unordered_map<int, vector<int>> mp;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> s;
  for (i = 1; i <= n; ++i) cin >> a[i];
  memcpy(aa + 1, a + 1, n << 2);
  sort(aa + 1, aa + n + 1);
  for (i = 1; i <= n; ++i) {
    bb[i] = a[i] == aa[i], g[i] = i;
    if (!bb[i]) S.insert(i);
  }
  int ss = S.size();
  if (ss > s) {
    cout << -1 << endl;
    return 0;
  }
  for (i = 1; i <= n; ++i) en[aa[i]] = i;
  for (i = n; i; --i) be[aa[i]] = i;
  for (i = 1; i <= n; ++i)
    if (!bb[i])
      p[i] = *S.lower_bound(be[a[i]]), S.erase(p[i]), g[gfa(p[i])] = gfa(i);
  for (i = 1; i <= n; ++i)
    if (!bb[i])
      mp[a[i]].push_back(i);
    else
      p[i] = i;
  for (auto u : mp) {
    auto v = u.second;
    for (i = 1; i < v.size(); ++i)
      if (gfa(v[i]) != gfa(v[0]))
        g[gfa(v[i])] = gfa(v[0]), swap(p[v[i]], p[v[0]]);
  }
  vector<vector<int>> ans;
  for (i = 1; i <= n; ++i)
    if (p[i] != i && !vi[i]) {
      vector<int> ve;
      for (j = i; vi[j] = 1, ve.push_back(j), p[j] != i; j = p[j])
        ;
      ans.push_back(ve);
    }
  if (ss + 3 <= s && ans.size() > 1) {
    vector<int> v1, v2, v3;
    for (i = 1; ss + i <= s && !ans.empty(); ++i) {
      v3 = ans.back();
      ans.pop_back();
      v1.insert(v1.end(), v3.begin(), v3.end());
      v2.push_back(v3[0]);
    }
    reverse(v2.begin(), v2.end());
    ans.push_back(v1);
    ans.push_back(v2);
  }
  cout << ans.size() << '\n';
  for (auto u : ans) {
    cout << u.size() << '\n';
    for (int x : u) cout << x << ' ';
    cout << '\n';
  }
  return 0;
}

```