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;
inline int read() {
  int n = 0;
  char c;
  for (c = getchar(); c < '0' || c > '9'; c = getchar())
    ;
  for (; c >= '0' && c <= '9'; c = getchar()) n = n * 10 + c - 48;
  return n;
}
const int maxn = 2e5 + 5;
int i, j, n, s, a[maxn], f[maxn], p[maxn], cnt, sum;
vector<int> an[maxn];
bool bz[maxn];
pair<int, int> b[maxn];
int get(int x) { return f[x] ? f[x] = get(f[x]) : x; }
void merge(int x, int y) {
  x = get(x), y = get(y);
  if (x != y) f[x] = y;
}
int main() {
  n = read(), s = read();
  for (i = 1; i <= n; i++) a[i] = read(), b[i] = make_pair(a[i], i);
  sort(b + 1, b + 1 + n);
  for (i = 1; i <= n; i++) p[b[i].second] = i;
  for (i = 1; i <= n; i++)
    if (a[i] == b[i].first && p[i] != i) {
      p[b[i].second] = p[i];
      b[p[i]].second = b[i].second;
      p[i] = i, b[i].second = i;
    }
  for (i = 1; i <= n; i++)
    if (p[i] != i) merge(p[i], i);
  int x = 0;
  for (i = 1; i <= n; i++) {
    int y = b[i].second;
    if (p[y] == y) continue;
    if (a[x] == a[y]) {
      if (get(x) == get(y)) continue;
      merge(x, y);
      swap(p[x], p[y]);
    }
    x = y;
  }
  for (i = 1; i <= n; i++)
    if (p[i] != i && !bz[i]) {
      cnt++, x = i;
      for (; !bz[x]; x = p[x]) sum++, bz[x] = 1, an[cnt].push_back(x);
    }
  if (s < sum) return puts("-1"), 0;
  s -= sum;
  s = min(s, cnt);
  if (s <= 2) {
    printf("%d\n", cnt);
    for (i = 1; i <= cnt; i++) {
      printf("%d\n", an[i].size());
      for (j = 0; j < an[i].size(); j++) printf("%d ", an[i][j]);
      puts("");
    }
    return 0;
  }
  printf("%d\n", cnt - s + 2);
  for (i = 1; i <= cnt - s; i++) {
    printf("%d\n", an[i].size());
    for (j = 0; j < an[i].size(); j++) printf("%d ", an[i][j]);
    sum -= an[i].size();
    puts("");
  }
  printf("%d\n", sum);
  for (i = cnt - s + 1; i <= cnt; i++)
    for (j = 0; j < an[i].size(); j++) printf("%d ", an[i][j]);
  puts("");
  printf("%d\n", s);
  for (i = cnt; i >= cnt - s + 1; i--) printf("%d ", an[i][0]);
  puts("");
}