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;
using uint = unsigned int;
using ll = long long;
using pii = pair<int, int>;
int n, s;
int a[200010], b[200010];
int p[200010];
bool used[200010];
int fth[200010];
vector<vector<int>> cycles;
map<int, vector<int>> v, pos;
int root(int x) { return fth[x] == -1 ? x : fth[x] = root(fth[x]); }
bool join(int x, int y) {
  x = root(x);
  y = root(y);
  if (x == y) return false;
  if ((x + y) & 1)
    fth[x] = y;
  else
    fth[y] = x;
  return true;
}
void getMinPerm() {
  int i;
  for (i = 1; i <= n; ++i) b[i] = a[i];
  sort(b + 1, b + n + 1);
  for (i = 1; i <= n; ++i)
    if (a[i] != b[i]) v[b[i]].push_back(i);
  for (i = 1; i <= n; ++i) {
    if (a[i] == b[i])
      p[i] = i;
    else {
      p[i] = v[a[i]].back();
      v[a[i]].pop_back();
    }
  }
  for (i = 1; i <= n; ++i) join(i, p[i]);
  for (i = 1; i <= n; ++i)
    if (a[i] != b[i]) pos[a[i]].push_back(i);
  for (const auto &item : pos) {
    for (auto val : item.second)
      if (join(p[val], p[item.second[0]])) {
        swap(p[val], p[item.second[0]]);
      }
  }
}
void solve() {
  int i, t;
  for (t = 0, i = 1; i <= n; ++i) {
    if (p[i] == i) continue;
    if (used[i]) continue;
    cycles.push_back(vector<int>());
    for (int j = i; !used[j]; used[j] = true, j = p[j])
      cycles.back().push_back(j);
    t += cycles.back().size();
  }
  int m = cycles.size();
  int x = max(m + t - s, 0);
  int y = m - x;
  if (t > s) return void(cout << "-1\n");
  cout << (x + min(y, 2)) << '\n';
  if (y == 1) ++x;
  while (x--) {
    cout << cycles.back().size() << '\n';
    for (auto val : cycles.back()) cout << val << ' ';
    cout << '\n';
    cycles.pop_back();
  }
  if (cycles.empty()) return;
  int sum = 0;
  for (i = cycles.size() - 1; i >= 0; --i) sum += cycles[i].size();
  cout << '\n';
  cout << sum << '\n';
  for (const auto &vec : cycles) {
    for (auto val : vec) cout << val << ' ';
  }
  cout << '\n';
  cout << cycles.size() << '\n';
  for (i = cycles.size() - 1; i >= 0; --i) cout << cycles[i][0] << ' ';
  cout << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  int i;
  memset(fth, -1, sizeof fth);
  cin >> n >> s;
  for (i = 1; i <= n; ++i) cin >> a[i];
  getMinPerm();
  solve();
  return 0;
}