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 MAXN = (int)2e5 + 3;
const int infint = (int)1e9 + 3;
const long long inf = (long long)1e17;
int n, s, a[MAXN], idx[MAXN], final[MAXN], par[MAXN], visited[MAXN];
set<int> id[MAXN], num[MAXN];
vector<int> cyc[MAXN];
bool cmp(int u, int v) { return a[u] < a[v]; }
int get(int u) { return par[u] < 0 ? u : par[u] = get(par[u]); }
void merge(int u, int v) {
  if ((u = get(u)) == (v = get(v))) return;
  if (par[u] > par[v]) swap(u, v);
  par[u] += par[v];
  par[u] = v;
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> s;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 1; i <= n; i++) idx[i] = i;
  sort(idx + 1, idx + n + 1, cmp);
  int sz = -1;
  for (int i = 1; i <= n; i++) {
    if (a[idx[i]] != a[idx[i - 1]]) sz++;
    id[sz].insert(i);
    num[sz].insert(idx[i]);
  }
  sz++;
  for (int i = 0; i < sz; i++) {
    vector<int> to_del;
    for (auto u : id[i])
      if (num[i].find(u) != num[i].end()) to_del.push_back(u);
    for (auto u : to_del) final[u] = u, num[i].erase(u), id[i].erase(u);
  }
  memset(par, -1, sizeof par);
  for (int i = 0; i < sz; i++) {
    vector<int> curid, curnum;
    for (auto u : id[i]) curid.push_back(u);
    for (auto u : num[i]) curnum.push_back(u);
    for (int j = 0; j < curnum.size(); j++)
      final[curnum[j]] = curid[j], merge(curid[j], curnum[j]);
  }
  for (int i = 0; i < sz; i++) {
    vector<int> curnum;
    for (auto u : num[i]) curnum.push_back(u);
    for (int j = 1; j < curnum.size(); j++)
      if (get(curnum[j]) != get(curnum[j - 1]))
        swap(final[curnum[j]], final[curnum[j - 1]]),
            merge(curnum[j], curnum[j - 1]);
  }
  sz = -1;
  for (int i = 1; i <= n; i++)
    if (final[i] != i && !visited[i]) {
      sz++;
      int iter = i;
      while (!visited[iter]) {
        cyc[sz].push_back(iter);
        visited[iter] = 1;
        iter = final[iter];
      }
    }
  sz++;
  int sum_of_cyc = 0;
  for (int i = 0; i < sz; i++) sum_of_cyc += cyc[i].size();
  s -= sum_of_cyc;
  if (s < 0) return cout << -1, 0;
  if (sz <= 2 || s <= 2) {
    cout << sz << "\n";
    for (int i = 0; i < sz; i++) {
      cout << cyc[i].size() << "\n";
      for (auto u : cyc[i]) cout << u << " ";
      cout << "\n";
    }
    return 0;
  }
  int merge_cyc = min(s, sz);
  cout << sz - merge_cyc + 2 << "\n";
  int s0 = 0;
  for (int i = 0; i < merge_cyc; i++) s0 += cyc[i].size();
  cout << s0 << "\n";
  for (int i = 0; i < merge_cyc; i++)
    for (auto u : cyc[i]) cout << u << " ";
  cout << "\n";
  cout << merge_cyc << "\n";
  cout << cyc[0][0] << " ";
  for (int i = merge_cyc - 1; i >= 1; i--) cout << cyc[i][0] << " ";
  cout << "\n";
  for (int i = merge_cyc; i < sz; i++) {
    cout << cyc[i].size() << "\n";
    for (auto u : cyc[i]) cout << u << " ";
    cout << "\n";
  }
}