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 = 200005;
int n, m, f[N], to[N];
pair<int, int> a[N];
bool done[N];
int find(int x) {
  while (x != f[x]) {
    x = f[x] = f[f[x]];
  }
  return x;
}
int main() {
  scanf("%d %d", &n, &m);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &a[i].first);
    a[i].second = i;
  }
  sort(a + 1, a + n + 1);
  for (int i = 1, j = 1; i <= n; i = j) {
    while (j <= n && a[j].first == a[i].first) {
      ++j;
    }
    for (int k = i; k < j; ++k) {
      if (a[k].second >= i && a[k].second < j) {
        to[a[k].second] = a[k].second;
        done[a[k].second] = true;
        ++m;
      }
    }
    int t = i;
    for (int k = i; k < j; ++k) {
      if (a[k].second < i || a[k].second >= j) {
        while (done[t]) {
          ++t;
        }
        to[a[k].second] = t++;
      }
    }
  }
  if (m < n) {
    puts("-1");
    return 0;
  }
  for (int i = 1; i <= n; ++i) {
    f[i] = i;
  }
  for (int i = 1; i <= n; ++i) {
    f[find(i)] = find(to[i]);
  }
  for (int i = 1, j = 1; i <= n; i = j) {
    while (j <= n && a[j].first == a[i].first) {
      ++j;
    }
    int last = 0;
    for (int k = i; k < j; ++k) {
      if (!done[a[k].second]) {
        int t = a[k].second;
        if (last && find(last) != find(t)) {
          f[find(last)] = find(t);
          swap(to[last], to[a[k].second]);
        }
        last = a[k].second;
      }
    }
  }
  int cycle = 0;
  for (int i = 1; i <= n; ++i) {
    if (find(i) == i && to[i] != i) {
      ++cycle;
    }
  }
  int change = min(cycle, m - n);
  vector<int> all, go;
  if (change > 2) {
    for (int i = 1; i <= n; ++i) {
      if (find(i) == i && to[i] != i) {
        all.push_back(i);
        go.push_back(to[i]);
        if (all.size() == change) {
          break;
        }
      }
    }
    for (int i = 0; i < change; ++i) {
      to[all[i]] = go[(i + change - 1) % change];
      if (i) {
        f[find(all[i])] = find(all[i - 1]);
      }
    }
  }
  int answer = 0;
  for (int i = 1; i <= n; ++i) {
    if (find(i) == i && to[i] != i) {
      ++answer;
    }
  }
  printf("%d\n", answer + (!go.empty()));
  for (int i = 1; i <= n; ++i) {
    if (find(i) == i && to[i] != i) {
      vector<int> a;
      a.push_back(i);
      for (int x = to[i]; x != i; x = to[x]) {
        a.push_back(x);
      }
      printf("%d\n", a.size());
      for (int j = 0; j < a.size(); ++j) {
        printf("%d%c", a[j], j == a.size() - 1 ? '\n' : ' ');
      }
    }
  }
  if (!go.empty()) {
    printf("%d\n", go.size());
    for (int i = 0; i < go.size(); ++i) {
      printf("%d%c", go[i], i == go.size() - 1 ? '\n' : ' ');
    }
  }
  return 0;
}