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 N = 2e5 + 10;
int a[N], b[N], c[N], d[N], e[N];
using VI = vector<int>;
VI v[N], rep[N];
int rn;
void search(int u) {
  for (int U; !v[u].empty();) {
    U = v[u].back();
    v[u].pop_back();
    search(a[U]);
    rep[rn].push_back(U);
  }
}
int main() {
  int n, s, m;
  scanf("%d %d", &n, &s);
  for (int i = 1; i <= n; i++) {
    scanf("%d", a + i);
    c[i] = a[i];
    d[i] = i;
  }
  sort(c + 1, c + n + 1);
  m = unique(c + 1, c + n + 1) - c - 1;
  for (int i = 1; i <= n; i++) {
    b[i] = a[i] = lower_bound(c + 1, c + m + 1, a[i]) - c;
  }
  sort(b + 1, b + n + 1);
  for (int i = 1; i <= n; i++) {
    if (a[i] != b[i]) {
      v[b[i]].push_back(i);
    }
  }
  for (int i = 1; i <= m; i++) {
    if (!v[i].empty()) {
      search(i);
      int f = rep[rn].size();
      for (int j = 0; j < f; j++) {
        d[rep[rn][j == f - 1 ? 0 : j + 1]] = rep[rn][j];
      }
      s -= f;
      rn++;
    }
    if (s < 0) {
      puts("-1");
      return 0;
    }
  }
  s = min(rn, s);
  if (s >= 2) {
    printf("%d\n", rn - s + 2);
    {
      printf("%d\n", s);
      for (int i = 0; i < s; i++) {
        printf("%d%c", rep[i][0], i + 1 < s ? ' ' : '\n');
      }
      a[N - 1] = d[rep[s - 1][0]];
      for (int i = s; --i;) {
        d[rep[i][0]] = d[rep[i - 1][0]];
      }
      d[rep[0][0]] = a[N - 1];
    }
  } else {
    printf("%d\n", rn);
  }
  VI g;
  for (int i = 1; i <= n; i++) {
    if (d[i] == i) continue;
    if (e[i]) continue;
    g.clear();
    for (int j = i; !e[j];) {
      g.push_back(j);
      e[j] = 1;
      j = d[j];
    }
    printf("%d\n", g.size());
    for (int j = 0; j < g.size(); j++) {
      printf("%d%c", g[j], j + 1 < g.size() ? ' ' : '\n');
    }
  }
  return 0;
}