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;
int n, s, a[200010];
map<int, int> mp1;
map<int, vector<int> > mp2;
vector<vector<int> > ans;
void dfs(int u) {
  while (!mp2[u].empty()) {
    int i = mp2[u].back();
    mp2[u].pop_back();
    dfs(a[i]);
    ans.back().push_back(i);
  }
  mp2.erase(u);
}
int main() {
  scanf("%d%d", &n, &s);
  for (int i = 1; i <= n; i++) scanf("%d", a + i), ++mp1[a[i]];
  int j = 1;
  for (auto it = mp1.begin(); it != mp1.end(); j += it->second, ++it)
    for (int i = j; i < j + it->second; i++)
      if (a[i] != it->first) mp2[it->first].push_back(i);
  while (!mp2.empty()) {
    ans.push_back(vector<int>());
    dfs(mp2.begin()->first);
    reverse(ans.back().begin(), ans.back().end());
    s -= ans.back().size();
  }
  if (s < 0) return puts("-1"), 0;
  int b = min(s, (int)ans.size()), d = 0;
  printf("%d\n", ans.size() - (b >= 3 ? b - 2 : 0));
  if (b >= 3) {
    for (int i = 0; i < b; i++) d += ans[i].size();
    printf("%d\n", d);
    for (int i = 0; i < b; i++)
      for (auto c : ans[i]) printf("%d ", c);
    printf("\n%d\n", b);
    for (int i = b - 1; i + 1; i--) printf("%d ", ans[i][0]);
    printf("\n");
  }
  for (int i = (b >= 3 ? b : 0); i < ans.size(); i++) {
    printf("%d\n", ans[i].size());
    for (auto c : ans[i]) printf("%d ", c);
    printf("\n");
  }
  return 0;
}