#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 5;
int a[N];
int b[N], t[N], nt[N], u;
vector<int> ans[N];
int top;
int NT(int x) {
  while (nt[x] < t[x + 1] && a[nt[x]] == x) ++nt[x];
  return nt[x];
}
void dfs(int x) {
  while (NT(x) < t[x + 1]) {
    int now = nt[x]++;
    dfs(a[now]);
    ans[top].push_back(now);
  }
}
int main() {
  int n, s;
  cin >> n >> s;
  for (int i = 1; i <= n; ++i) scanf("%d", a + i);
  for (int i = 1; i <= n; ++i) b[i] = a[i];
  sort(b + 1, b + n + 1);
  for (int i = 1; i <= n; ++i)
    if (b[i] != b[u]) {
      ++u;
      t[u] = i;
      b[u] = b[i];
    }
  t[u + 1] = n + 1;
  for (int i = 1; i <= n; ++i) a[i] = lower_bound(b + 1, b + u + 1, a[i]) - b;
  for (int i = 1; i <= u; ++i) nt[i] = t[i];
  for (int i = 1; i <= u; ++i)
    if (NT(i) < t[i + 1]) {
      ++top;
      dfs(i);
      s -= ans[top].size();
      reverse(ans[top].begin(), ans[top].end());
    }
  if (s < 0)
    puts("-1");
  else {
    s = min(s, top);
    if (s < 2) {
      printf("%d\n", top);
      for (int i = 1; i <= top; ++i) {
        printf("%d\n", (int)ans[i].size());
        for (auto x : ans[i]) printf("%d ", x);
        puts("");
      }
    } else {
      printf("%d\n", top - s + 2);
      int sum = 0;
      for (int i = 1; i <= s; ++i) sum += ans[i].size();
      printf("%d\n", sum);
      for (int i = 1; i <= s; ++i)
        for (auto x : ans[i]) printf("%d ", x);
      puts("");
      printf("%d\n", s);
      for (int i = s; i >= 1; --i) printf("%d ", ans[i][0]);
      puts("");
      for (int i = s + 1; i <= top; ++i) {
        printf("%d\n", (int)ans[i].size());
        for (auto x : ans[i]) printf("%d ", x);
        puts("");
      }
    }
  }
}

### Prompt

In cpp, your task is to solve 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 a[N];
int b[N], t[N], nt[N], u;
vector<int> ans[N];
int top;
int NT(int x) {
  while (nt[x] < t[x + 1] && a[nt[x]] == x) ++nt[x];
  return nt[x];
}
void dfs(int x) {
  while (NT(x) < t[x + 1]) {
    int now = nt[x]++;
    dfs(a[now]);
    ans[top].push_back(now);
  }
}
int main() {
  int n, s;
  cin >> n >> s;
  for (int i = 1; i <= n; ++i) scanf("%d", a + i);
  for (int i = 1; i <= n; ++i) b[i] = a[i];
  sort(b + 1, b + n + 1);
  for (int i = 1; i <= n; ++i)
    if (b[i] != b[u]) {
      ++u;
      t[u] = i;
      b[u] = b[i];
    }
  t[u + 1] = n + 1;
  for (int i = 1; i <= n; ++i) a[i] = lower_bound(b + 1, b + u + 1, a[i]) - b;
  for (int i = 1; i <= u; ++i) nt[i] = t[i];
  for (int i = 1; i <= u; ++i)
    if (NT(i) < t[i + 1]) {
      ++top;
      dfs(i);
      s -= ans[top].size();
      reverse(ans[top].begin(), ans[top].end());
    }
  if (s < 0)
    puts("-1");
  else {
    s = min(s, top);
    if (s < 2) {
      printf("%d\n", top);
      for (int i = 1; i <= top; ++i) {
        printf("%d\n", (int)ans[i].size());
        for (auto x : ans[i]) printf("%d ", x);
        puts("");
      }
    } else {
      printf("%d\n", top - s + 2);
      int sum = 0;
      for (int i = 1; i <= s; ++i) sum += ans[i].size();
      printf("%d\n", sum);
      for (int i = 1; i <= s; ++i)
        for (auto x : ans[i]) printf("%d ", x);
      puts("");
      printf("%d\n", s);
      for (int i = s; i >= 1; --i) printf("%d ", ans[i][0]);
      puts("");
      for (int i = s + 1; i <= top; ++i) {
        printf("%d\n", (int)ans[i].size());
        for (auto x : ans[i]) printf("%d ", x);
        puts("");
      }
    }
  }
}

```