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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int maxN = 202000;
int n, S, ufs[maxN], Id[maxN], vis[maxN];
pair<int, int> A[maxN];
vector<int> Ring[maxN];
int find(int x);
void dfs(int rcnt, int u);
int main() {
  scanf("%d%d", &n, &S);
  for (int i = 1; i <= n; i++)
    scanf("%d", &A[i].first), A[i].second = ufs[i] = i;
  sort(&A[1], &A[n + 1]);
  for (int i = 1; i <= n; i++) Id[A[i].second] = i;
  for (int i = 1; i <= n; i++)
    if (A[i].first == A[Id[i]].first && i != Id[i]) {
      int x = A[i].second, y = Id[i];
      A[y].second = x;
      Id[x] = y;
      Id[i] = A[i].second = i;
    }
  int sum = 0, rcnt = 0;
  for (int i = 1; i <= n; i++) ufs[find(i)] = find(Id[i]);
  for (int i = 1, lst = 0; i <= n; i++)
    if (Id[i] != i) {
      ++sum;
      if (lst && A[lst].first == A[i].first &&
          find(A[lst].second) != find(A[i].second)) {
        ufs[find(A[lst].second)] = find(A[i].second);
        swap(Id[A[lst].second], Id[A[i].second]);
      }
      lst = i;
    }
  for (int i = 1; i <= n; i++)
    if (Id[i] != i && !vis[i]) dfs(++rcnt, i);
  if (sum > S) {
    puts("-1");
    return 0;
  }
  int trn = min(S - sum, rcnt);
  if (trn <= 2) {
    printf("%d\n", rcnt);
    for (int i = 1, sz; i <= rcnt; i++) {
      printf("%d\n", sz = Ring[i].size());
      for (int j = 0; j < sz; j++) printf("%d ", Ring[i][j]);
      printf("\n");
    }
  } else {
    printf("%d\n", rcnt - (trn - 2));
    for (int i = 1, sz; i <= rcnt - trn; i++) {
      printf("%d\n", sz = Ring[i].size());
      for (int j = 0; j < sz; j++) printf("%d ", Ring[i][j]);
      printf("\n");
    }
    int sum = 0;
    for (int i = rcnt - trn + 1; i <= rcnt; i++) sum += Ring[i].size();
    printf("%d\n", sum);
    for (int i = rcnt - trn + 1; i <= rcnt; i++)
      for (int j = 0, sz = Ring[i].size(); j < sz; j++)
        printf("%d ", Ring[i][j]);
    printf("\n");
    printf("%d\n", trn);
    for (int i = rcnt; i >= rcnt - trn + 1; i--) printf("%d ", Ring[i][0]);
    printf("\n");
  }
  return 0;
}
int find(int x) {
  if (ufs[x] != x) ufs[x] = find(ufs[x]);
  return ufs[x];
}
void dfs(int rcnt, int u) {
  if (vis[u]) return;
  Ring[rcnt].push_back(u);
  vis[u] = 1;
  dfs(rcnt, Id[u]);
  return;
}