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;
template <class T, class U>
void ckmin(T &a, U b) {
  if (a > b) a = b;
}
template <class T, class U>
void ckmax(T &a, U b) {
  if (a < b) a = b;
}
const int MAXN = 400013;
int N, S, M, ans, n;
int val[MAXN], arr[MAXN], sorted[MAXN];
vector<int> moves[MAXN];
vector<int> cyc[MAXN];
bitset<MAXN> vis;
vector<int> compress;
int freq[MAXN];
pair<int, int> range[MAXN];
vector<int> edge[MAXN];
vector<int> tour;
int ind[MAXN];
int indexof(vector<int> &v, int x) {
  return upper_bound((v).begin(), (v).end(), x) - v.begin() - 1;
}
void dfs(int u) {
  while (ind[u] < ((int)(edge[u]).size())) {
    dfs(edge[u][ind[u]++]);
  }
  tour.push_back(u);
}
void solve() {
  int k = min(M, S - n);
  if (k <= 2) {
    ans = M;
    for (auto i = (0); i < (M); i++) {
      moves[i] = cyc[i];
    }
  } else {
    ans = M - k + 2;
    for (auto i = (0); i < (M - k); i++) {
      moves[i] = cyc[i];
    }
    for (auto i = (M - k); i < (M); i++) {
      moves[M - k].insert(moves[M - k].end(), (cyc[i]).begin(), (cyc[i]).end());
      moves[M - k + 1].push_back(cyc[i][0]);
    }
    reverse((moves[M - k + 1]).begin(), (moves[M - k + 1]).end());
  }
}
int32_t main() {
  cout << fixed << setprecision(12);
  cerr << fixed << setprecision(4);
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> N >> S;
  for (auto i = (0); i < (N); i++) {
    cin >> val[i];
    compress.push_back(val[i]);
  }
  sort((compress).begin(), (compress).end());
  compress.erase(unique((compress).begin(), (compress).end()), compress.end());
  for (auto i = (0); i < (N); i++) {
    val[i] = indexof(compress, val[i]);
    sorted[i] = val[i];
  }
  sort(sorted, sorted + N);
  for (auto i = (0); i < (N); i++) {
    if (val[i] == sorted[i]) {
      vis[i] = true;
      arr[i] = i;
      continue;
    }
    edge[i].push_back(val[i] + N);
    edge[sorted[i] + N].push_back(i);
  }
  for (auto i = (0); i < (N); i++) {
    if (vis[i]) continue;
    tour.clear();
    dfs(i);
    reverse((tour).begin(), (tour).end());
    for (int j = 0; j + 2 < ((int)(tour).size()); j += 2) {
      int u = tour[j];
      int v = tour[j + 2];
      arr[u] = v;
    }
    tour.clear();
  }
  vis.reset();
  for (auto i = (0); i < (N); i++) {
    if (vis[i]) continue;
    if (arr[i] == i) {
      continue;
    }
    cyc[M].push_back(i);
    do {
      vis[cyc[M].back()] = true;
      cyc[M].push_back(arr[cyc[M].back()]);
    } while (cyc[M].back() != i);
    cyc[M].pop_back();
    M++;
  }
  for (auto i = (0); i < (M); i++) {
    n += ((int)(cyc[i]).size());
  }
  if (S < n) {
    cout << "-1\n";
    return 0;
  }
  solve();
  assert(ans == min(M, max(2, 2 + M - S + n)));
  cout << ans << '\n';
  for (auto i = (0); i < (ans); i++) {
    cout << ((int)(moves[i]).size()) << '\n';
    for (int x : moves[i]) {
      cout << x + 1 << " \n"[x == moves[i].back()];
    }
  }
}