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 = 1000005;
int n, a[N], K, sum;
map<int, int> mp;
map<int, vector<int> > g;
vector<vector<int> > ans;
vector<int> vec;
inline void dfs(const int x) {
  register int i;
  while (!g[x].empty())
    i = g[x].back(), g[x].pop_back(), dfs(a[i]), vec.push_back(i);
  g.erase(x);
}
int main() {
  scanf("%d%d", &n, &K);
  register int i;
  for (i = 1; i <= n; ++i) scanf("%d", &a[i]), ++mp[a[i]];
  for (const auto &x : mp)
    for (i = sum + 1, sum += x.second; i <= sum; ++i)
      if (a[i] ^ x.first) g[x.first].push_back(i);
  while (!g.empty())
    vec.clear(), dfs((*g.begin()).first), std::reverse(vec.begin(), vec.end()),
        ans.push_back(vec), K -= vec.size();
  if (K < 0) return puts("-1"), 0;
  ans.size() < K ? K = ans.size() : 0,
                   printf("%d\n", ans.size() - (K > 2 ? K - 2 : 0));
  if (K < 3) {
    for (const auto &o : ans) {
      printf("%d\n", o.size());
      for (const int &i : o) printf("%d ", i);
      puts("");
    }
    return 0;
  }
  register int len = 0;
  for (i = 0; i < K; ++i) len += ans[i].size();
  printf("%d\n", len);
  for (i = 0; i < K; ++i)
    for (const int &j : ans[i]) printf("%d ", j);
  printf("\n%d\n", K);
  for (i = K - 1; ~i; --i) printf("%d ", ans[i][0]);
  puts("");
  for (i = K; i < ans.size(); ++i) {
    printf("%d\n", ans[i].size());
    for (const int &j : ans[i]) printf("%d ", j);
    puts("");
  }
  return 0;
}