#include <bits/stdc++.h>
using namespace std;
int n, s, a[200005], b[200005], c[200005], vis[200005], wa, cir, prn[200005],
    to[200005], nex[200005], beg[200005], val[200005], e;
bool acs[200005];
void insert(int x, int y, int z) {
  to[++e] = y;
  nex[e] = beg[x];
  beg[x] = e;
  val[e] = z;
}
vector<int> tek[200005];
void dfs(int now) {
  vis[now] = cir;
  for (int &i = beg[now]; i; i = nex[i]) {
    if (!acs[i]) {
      int tmp = i;
      acs[i] = true;
      dfs(to[i]);
      tek[cir].push_back(val[tmp]);
    }
  }
}
int main() {
  scanf("%d %d", &n, &s);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]), b[i] = a[i];
  sort(b + 1, b + 1 + n);
  int len = unique(b + 1, b + 1 + n) - b - 1;
  for (int i = 1; i <= n; ++i)
    c[i] = a[i] = lower_bound(b + 1, b + 1 + len, a[i]) - b;
  sort(c + 1, c + 1 + n);
  for (int i = 1; i <= n; ++i)
    if (c[i] != a[i]) ++wa, insert(a[i], c[i], i);
  if (wa > s) return puts("-1") & 0;
  for (int i = 1; i <= n; ++i)
    if (!vis[i] && beg[i]) ++cir, dfs(i);
  if (cir <= 1 || s - wa <= 1) {
    printf("%d\n", cir);
    for (int i = 1; i <= cir; ++i) {
      printf("%d\n", int(tek[i].size()));
      for (unsigned int j = 0; j < tek[i].size(); ++j) printf("%d ", tek[i][j]);
      puts("");
    }
  } else {
    printf("%d\n", cir - min(cir, s - wa) + 2);
    for (int i = s - wa + 1; i <= cir; ++i) {
      printf("%d\n", int(tek[i].size()));
      for (unsigned int j = 0; j < tek[i].size(); ++j) printf("%d ", tek[i][j]);
      puts("");
    }
    if (s != wa) {
      int p1 = 0, p2 = 0;
      for (int i = min(s - wa, cir); i; --i)
        p1 += int(tek[i].size()), prn[++p2] = tek[i][0];
      printf("%d\n", p1);
      for (int i = 1; i <= min(s - wa, cir); ++i)
        for (unsigned int j = 0; j < tek[i].size(); ++j)
          printf("%d ", tek[i][j]);
      puts("");
      printf("%d\n", p2);
      for (int i = 1; i <= p2; ++i) printf("%d ", prn[i]);
    }
  }
  return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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;
int n, s, a[200005], b[200005], c[200005], vis[200005], wa, cir, prn[200005],
    to[200005], nex[200005], beg[200005], val[200005], e;
bool acs[200005];
void insert(int x, int y, int z) {
  to[++e] = y;
  nex[e] = beg[x];
  beg[x] = e;
  val[e] = z;
}
vector<int> tek[200005];
void dfs(int now) {
  vis[now] = cir;
  for (int &i = beg[now]; i; i = nex[i]) {
    if (!acs[i]) {
      int tmp = i;
      acs[i] = true;
      dfs(to[i]);
      tek[cir].push_back(val[tmp]);
    }
  }
}
int main() {
  scanf("%d %d", &n, &s);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]), b[i] = a[i];
  sort(b + 1, b + 1 + n);
  int len = unique(b + 1, b + 1 + n) - b - 1;
  for (int i = 1; i <= n; ++i)
    c[i] = a[i] = lower_bound(b + 1, b + 1 + len, a[i]) - b;
  sort(c + 1, c + 1 + n);
  for (int i = 1; i <= n; ++i)
    if (c[i] != a[i]) ++wa, insert(a[i], c[i], i);
  if (wa > s) return puts("-1") & 0;
  for (int i = 1; i <= n; ++i)
    if (!vis[i] && beg[i]) ++cir, dfs(i);
  if (cir <= 1 || s - wa <= 1) {
    printf("%d\n", cir);
    for (int i = 1; i <= cir; ++i) {
      printf("%d\n", int(tek[i].size()));
      for (unsigned int j = 0; j < tek[i].size(); ++j) printf("%d ", tek[i][j]);
      puts("");
    }
  } else {
    printf("%d\n", cir - min(cir, s - wa) + 2);
    for (int i = s - wa + 1; i <= cir; ++i) {
      printf("%d\n", int(tek[i].size()));
      for (unsigned int j = 0; j < tek[i].size(); ++j) printf("%d ", tek[i][j]);
      puts("");
    }
    if (s != wa) {
      int p1 = 0, p2 = 0;
      for (int i = min(s - wa, cir); i; --i)
        p1 += int(tek[i].size()), prn[++p2] = tek[i][0];
      printf("%d\n", p1);
      for (int i = 1; i <= min(s - wa, cir); ++i)
        for (unsigned int j = 0; j < tek[i].size(); ++j)
          printf("%d ", tek[i][j]);
      puts("");
      printf("%d\n", p2);
      for (int i = 1; i <= p2; ++i) printf("%d ", prn[i]);
    }
  }
  return 0;
}

```