#include <bits/stdc++.h>
using namespace std;
inline int read() {
  int x = 0, f = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = (x << 1) + (x << 3) + c - '0';
    c = getchar();
  }
  return x * f;
}
int n, m, len, a[200020], b[200020], c[200020], cnt, need, vis[200020], qwq;
vector<pair<int, int> > G[200020];
vector<int> ans[200020];
void dfs(int u) {
  vis[u] = 1;
  while (!G[u].empty()) {
    int v = G[u].back().first, w = G[u].back().second;
    G[u].pop_back();
    dfs(v);
    ans[cnt].push_back(w);
  }
}
int main() {
  n = read(), m = read();
  for (int i = 1; i <= n; ++i) {
    a[i] = c[i] = read();
  }
  sort(c + 1, c + n + 1);
  for (int i = 1; i <= n; ++i) {
    if (c[i] ^ c[i - 1]) ++len;
    b[i] = len;
  }
  len = unique(c + 1, c + n + 1) - c - 1;
  for (int i = 1; i <= n; ++i) {
    a[i] = lower_bound(c + 1, c + len + 1, a[i]) - c;
  }
  for (int i = 1; i <= n; ++i) {
    if (a[i] ^ b[i]) G[a[i]].push_back(make_pair(b[i], i)), ++need;
  }
  if (need > m) {
    return !printf("-1\n");
  }
  if (!need) {
    return !printf("0\n");
  }
  for (int i = 1; i <= len; ++i) {
    if (!vis[i] && !G[i].empty()) {
      ++cnt, dfs(i);
    }
  }
  if (cnt <= 1 || m - need <= 1) {
    printf("%d\n", cnt);
    for (int i = 1; i <= cnt; ++i) {
      printf("%d\n", (int)ans[i].size());
      for (auto x : ans[i]) {
        printf("%d ", x);
      }
      printf("\n");
    }
    return 0;
  }
  printf("%d\n", qwq = cnt + 2 - min(m - need, cnt));
  printf("%d\n", min(m - need, cnt));
  int tot = 0;
  for (int i = 1; i <= min(m - need, cnt); ++i) {
    printf("%d ", ans[i].back()), tot += ans[i].size();
  }
  printf("\n");
  reverse(ans + 1, ans + min(m - need, cnt) + 1);
  printf("%d\n", tot);
  for (int i = 1; i <= min(m - need, cnt); ++i) {
    for (auto x : ans[i]) {
      printf("%d ", x);
    }
  }
  printf("\n");
  for (int i = min(m - need, cnt) + 1; i <= cnt; ++i) {
    printf("%d\n", (int)ans[i].size());
    for (auto x : ans[i]) {
      printf("%d ", x);
    }
    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;
inline int read() {
  int x = 0, f = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = (x << 1) + (x << 3) + c - '0';
    c = getchar();
  }
  return x * f;
}
int n, m, len, a[200020], b[200020], c[200020], cnt, need, vis[200020], qwq;
vector<pair<int, int> > G[200020];
vector<int> ans[200020];
void dfs(int u) {
  vis[u] = 1;
  while (!G[u].empty()) {
    int v = G[u].back().first, w = G[u].back().second;
    G[u].pop_back();
    dfs(v);
    ans[cnt].push_back(w);
  }
}
int main() {
  n = read(), m = read();
  for (int i = 1; i <= n; ++i) {
    a[i] = c[i] = read();
  }
  sort(c + 1, c + n + 1);
  for (int i = 1; i <= n; ++i) {
    if (c[i] ^ c[i - 1]) ++len;
    b[i] = len;
  }
  len = unique(c + 1, c + n + 1) - c - 1;
  for (int i = 1; i <= n; ++i) {
    a[i] = lower_bound(c + 1, c + len + 1, a[i]) - c;
  }
  for (int i = 1; i <= n; ++i) {
    if (a[i] ^ b[i]) G[a[i]].push_back(make_pair(b[i], i)), ++need;
  }
  if (need > m) {
    return !printf("-1\n");
  }
  if (!need) {
    return !printf("0\n");
  }
  for (int i = 1; i <= len; ++i) {
    if (!vis[i] && !G[i].empty()) {
      ++cnt, dfs(i);
    }
  }
  if (cnt <= 1 || m - need <= 1) {
    printf("%d\n", cnt);
    for (int i = 1; i <= cnt; ++i) {
      printf("%d\n", (int)ans[i].size());
      for (auto x : ans[i]) {
        printf("%d ", x);
      }
      printf("\n");
    }
    return 0;
  }
  printf("%d\n", qwq = cnt + 2 - min(m - need, cnt));
  printf("%d\n", min(m - need, cnt));
  int tot = 0;
  for (int i = 1; i <= min(m - need, cnt); ++i) {
    printf("%d ", ans[i].back()), tot += ans[i].size();
  }
  printf("\n");
  reverse(ans + 1, ans + min(m - need, cnt) + 1);
  printf("%d\n", tot);
  for (int i = 1; i <= min(m - need, cnt); ++i) {
    for (auto x : ans[i]) {
      printf("%d ", x);
    }
  }
  printf("\n");
  for (int i = min(m - need, cnt) + 1; i <= cnt; ++i) {
    printf("%d\n", (int)ans[i].size());
    for (auto x : ans[i]) {
      printf("%d ", x);
    }
    printf("\n");
  }
  return 0;
}

```