#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 10;
int n, s;
int niz[maxn], sol[maxn], saz[maxn];
vector<pair<int, int> > graph[maxn];
bool bio[maxn], bio2[maxn];
vector<int> sa;
vector<vector<int> > al;
vector<int> ac[maxn];
void dfs(int node) {
  bio2[node] = true;
  while (!graph[node].empty()) {
    const int nig = graph[node].back().first;
    const int id = graph[node].back().second;
    graph[node].pop_back();
    dfs(nig);
    sa.push_back(id + 1);
  }
}
int main() {
  memset(bio, false, sizeof bio);
  memset(bio2, false, sizeof bio2);
  scanf("%d%d", &n, &s);
  for (int i = 0; i < n; i++) scanf("%d", niz + i);
  for (int i = 0; i < n; i++) sol[i] = niz[i];
  sort(sol, sol + n);
  for (int i = 0; i < n; i++) saz[i] = sol[i];
  for (int i = 0; i < n; i++)
    niz[i] = lower_bound(saz, saz + n, niz[i]) - saz,
    sol[i] = lower_bound(saz, saz + n, sol[i]) - saz;
  for (int i = 0; i < n; i++) {
    if (niz[i] == sol[i]) continue;
    graph[sol[i]].push_back(make_pair(niz[i], i));
    s--;
  }
  if (s < 0) {
    printf("-1");
    return 0;
  }
  for (int i = 0; i < n; i++)
    if (niz[i] != sol[i]) ac[niz[i]].push_back(i + 1);
  for (int i = 0; i < n; i++) {
    if (bio2[i]) continue;
    dfs(i);
    reverse(sa.begin(), sa.end());
    if (sa.size() == 0) continue;
    al.push_back(sa);
    sa.clear();
  }
  if (s > 2 && al.size() > 1) {
    int ptr = 0;
    vector<int> pok, ne;
    int siz = (int)al.size();
    for (int i = 0; i < min(s, siz); i++) {
      for (int j = 0; j < al.back().size(); j++) {
        if (j == 0) pok.push_back(al.back()[j]);
        ne.push_back(al.back()[j]);
      }
      al.pop_back();
    }
    al.push_back(ne);
    reverse(pok.begin(), pok.end());
    al.push_back(pok);
  }
  printf("%d\n", al.size());
  for (int i = 0; i < al.size(); i++) {
    printf("%d\n", al[i].size());
    for (int j = 0; j < al[i].size(); j++) printf("%d ", al[i][j]);
    printf("\n");
  }
  return 0;
}

### Prompt

Your challenge is to write 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 maxn = 2e5 + 10;
int n, s;
int niz[maxn], sol[maxn], saz[maxn];
vector<pair<int, int> > graph[maxn];
bool bio[maxn], bio2[maxn];
vector<int> sa;
vector<vector<int> > al;
vector<int> ac[maxn];
void dfs(int node) {
  bio2[node] = true;
  while (!graph[node].empty()) {
    const int nig = graph[node].back().first;
    const int id = graph[node].back().second;
    graph[node].pop_back();
    dfs(nig);
    sa.push_back(id + 1);
  }
}
int main() {
  memset(bio, false, sizeof bio);
  memset(bio2, false, sizeof bio2);
  scanf("%d%d", &n, &s);
  for (int i = 0; i < n; i++) scanf("%d", niz + i);
  for (int i = 0; i < n; i++) sol[i] = niz[i];
  sort(sol, sol + n);
  for (int i = 0; i < n; i++) saz[i] = sol[i];
  for (int i = 0; i < n; i++)
    niz[i] = lower_bound(saz, saz + n, niz[i]) - saz,
    sol[i] = lower_bound(saz, saz + n, sol[i]) - saz;
  for (int i = 0; i < n; i++) {
    if (niz[i] == sol[i]) continue;
    graph[sol[i]].push_back(make_pair(niz[i], i));
    s--;
  }
  if (s < 0) {
    printf("-1");
    return 0;
  }
  for (int i = 0; i < n; i++)
    if (niz[i] != sol[i]) ac[niz[i]].push_back(i + 1);
  for (int i = 0; i < n; i++) {
    if (bio2[i]) continue;
    dfs(i);
    reverse(sa.begin(), sa.end());
    if (sa.size() == 0) continue;
    al.push_back(sa);
    sa.clear();
  }
  if (s > 2 && al.size() > 1) {
    int ptr = 0;
    vector<int> pok, ne;
    int siz = (int)al.size();
    for (int i = 0; i < min(s, siz); i++) {
      for (int j = 0; j < al.back().size(); j++) {
        if (j == 0) pok.push_back(al.back()[j]);
        ne.push_back(al.back()[j]);
      }
      al.pop_back();
    }
    al.push_back(ne);
    reverse(pok.begin(), pok.end());
    al.push_back(pok);
  }
  printf("%d\n", al.size());
  for (int i = 0; i < al.size(); i++) {
    printf("%d\n", al[i].size());
    for (int j = 0; j < al[i].size(); j++) printf("%d ", al[i][j]);
    printf("\n");
  }
  return 0;
}

```