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 = 2e5 + 100;
int n, limit;
int nxt[N], lab[N], flag[N];
pair<int, int> a[N];
vector<int> ans1, ans2;
vector<vector<int> > ans3;
vector<pair<int, int> > v;
int root(int u) {
  if (lab[u] < 0) return u;
  return lab[u] = root(lab[u]);
}
void join(int u, int v) {
  int l1 = root(u), l2 = root(v);
  if (l1 == l2) return;
  if (lab[l1] > lab[l2]) swap(l1, l2);
  lab[l1] += lab[l2];
  lab[l2] = l1;
}
void show(vector<int> &s) {
  cout << s.size() << '\n';
  for (auto &x : s) cout << x << ' ';
  cout << '\n';
}
int main() {
  ios::sync_with_stdio(0);
  cin >> n >> limit;
  memset(lab, -1, sizeof lab);
  for (int i = 1; i <= n; ++i) {
    cin >> a[i].first;
    a[i].second = i;
  }
  sort(a + 1, a + n + 1);
  for (int i = 1; i <= n; ++i) {
    int j1 = i, j2 = i;
    while (j2 < n && a[j2 + 1].first == a[j2].first) j2++;
    for (; i <= j2; ++i) {
      while (j1 <= a[i].second && a[i].second <= j2 && a[i].second != i)
        swap(a[i], a[a[i].second]);
    }
    i = j2;
  }
  for (int i = 1; i <= n; ++i)
    if (a[i].second != i) {
      nxt[a[i].second] = i;
      join(a[i].second, i);
      v.push_back(a[i]);
    }
  for (int i = 0; i < v.size(); ++i) {
    while (i + 1 < v.size() && v[i + 1].first == v[i].first) {
      i++;
      int idx1 = v[i].second, idx2 = v[i - 1].second;
      if (root(idx1) == root(idx2)) continue;
      join(idx1, idx2);
      swap(nxt[idx1], nxt[idx2]);
    }
  }
  int sum = 0, tmp = 0;
  for (int i = 1; i <= n; ++i)
    if (a[i].second != i && !flag[root(a[i].second)]) {
      flag[root(a[i].second)] = true;
      tmp++;
      sum += abs(lab[root(a[i].second)]);
    }
  if (limit < sum) {
    cout << -1;
    return 0;
  }
  int addmx = limit - sum;
  int cnt = 0;
  memset(flag, 0, sizeof flag);
  for (int i = 1; i <= n; ++i)
    if (a[i].second != i && !flag[root(a[i].second)]) {
      flag[root(a[i].second)] = true;
      cnt++;
      if (cnt <= addmx) {
        ans2.push_back(a[i].second);
        int cur = a[i].second;
        do {
          ans1.push_back(cur);
          cur = nxt[cur];
        } while (cur != a[i].second);
      } else {
        vector<int> cycle;
        int cur = a[i].second;
        do {
          cycle.push_back(cur);
          cur = nxt[cur];
        } while (cur != a[i].second);
        ans3.push_back(cycle);
      }
    }
  if (ans1.size()) ans3.push_back(ans1);
  reverse(ans2.begin(), ans2.end());
  if (ans2.size() > 1) ans3.push_back(ans2);
  cout << ans3.size() << '\n';
  for (auto &s : ans3) show(s);
  return 0;
}