#include <bits/stdc++.h>
using namespace std;
using VI = vector<int>;
const int NN = 200011;
int a[NN], b[NN], arr[NN];
VI vec[NN], cyc[NN];
int nc;
int dp[NN], vst[NN];
int n, s;
void dfs(int u) {
  while (not vec[u].empty()) {
    int v = vec[u].back();
    vec[u].pop_back();
    dfs(a[v]);
    cyc[nc].push_back(v);
  }
}
int solve() {
  cin >> n >> s;
  for (int i = 1; i <= n; i++) scanf("%d", a + i), arr[i] = a[i], dp[i] = i;
  sort(arr + 1, arr + n + 1);
  int m = unique(arr + 1, arr + n + 1) - arr - 1;
  for (int i = 1; i <= n; i++) {
    b[i] = a[i] = lower_bound(arr + 1, arr + m + 1, a[i]) - arr;
  }
  sort(b + 1, b + n + 1);
  for (int i = 1; i <= n; i++)
    if (a[i] ^ b[i]) {
      vec[b[i]].push_back(i);
    }
  for (int i = 1; i <= m; i++) {
    if (not vec[i].empty()) {
      dfs(i);
      int r = ((int)cyc[nc].size());
      for (int j = 0; j < r; j++) {
        dp[cyc[nc][(j + 1) % r]] = cyc[nc][j];
        s--;
      }
      nc++;
    }
    if (s < 0) return puts("-1");
  }
  s = min(s, nc);
  if (s >= 2) {
    cout << nc - s + 2 << endl;
    cout << s << endl;
    for (int i = 0; i < s; i++) printf("%d ", cyc[i][0]);
    puts("");
    int tmp = dp[cyc[s - 1][0]];
    for (int i = s; --i;) {
      dp[cyc[i][0]] = dp[cyc[i - 1][0]];
    }
    dp[cyc[0][0]] = tmp;
  } else
    cout << nc << endl;
  for (int i = 1; i <= n; i++) {
    if (dp[i] == i or vst[i]) continue;
    VI ans;
    for (int u = i; not vst[u];) {
      ans.push_back(u);
      vst[u] = 1;
      u = dp[u];
    }
    printf("%d\n", ans.size());
    for (int u : ans) printf("%d ", u);
    puts("");
  }
}
int main() { solve(); }

### Prompt

Please provide a CPP coded solution 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;
using VI = vector<int>;
const int NN = 200011;
int a[NN], b[NN], arr[NN];
VI vec[NN], cyc[NN];
int nc;
int dp[NN], vst[NN];
int n, s;
void dfs(int u) {
  while (not vec[u].empty()) {
    int v = vec[u].back();
    vec[u].pop_back();
    dfs(a[v]);
    cyc[nc].push_back(v);
  }
}
int solve() {
  cin >> n >> s;
  for (int i = 1; i <= n; i++) scanf("%d", a + i), arr[i] = a[i], dp[i] = i;
  sort(arr + 1, arr + n + 1);
  int m = unique(arr + 1, arr + n + 1) - arr - 1;
  for (int i = 1; i <= n; i++) {
    b[i] = a[i] = lower_bound(arr + 1, arr + m + 1, a[i]) - arr;
  }
  sort(b + 1, b + n + 1);
  for (int i = 1; i <= n; i++)
    if (a[i] ^ b[i]) {
      vec[b[i]].push_back(i);
    }
  for (int i = 1; i <= m; i++) {
    if (not vec[i].empty()) {
      dfs(i);
      int r = ((int)cyc[nc].size());
      for (int j = 0; j < r; j++) {
        dp[cyc[nc][(j + 1) % r]] = cyc[nc][j];
        s--;
      }
      nc++;
    }
    if (s < 0) return puts("-1");
  }
  s = min(s, nc);
  if (s >= 2) {
    cout << nc - s + 2 << endl;
    cout << s << endl;
    for (int i = 0; i < s; i++) printf("%d ", cyc[i][0]);
    puts("");
    int tmp = dp[cyc[s - 1][0]];
    for (int i = s; --i;) {
      dp[cyc[i][0]] = dp[cyc[i - 1][0]];
    }
    dp[cyc[0][0]] = tmp;
  } else
    cout << nc << endl;
  for (int i = 1; i <= n; i++) {
    if (dp[i] == i or vst[i]) continue;
    VI ans;
    for (int u = i; not vst[u];) {
      ans.push_back(u);
      vst[u] = 1;
      u = dp[u];
    }
    printf("%d\n", ans.size());
    for (int u : ans) printf("%d ", u);
    puts("");
  }
}
int main() { solve(); }

```