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 = 2e5 + 5;
int a[maxn], n, s, b[maxn], num[maxn], len;
int ufs[maxn], meme[maxn];
inline int find(int u) {
  if (ufs[u] == u)
    return u;
  else
    return ufs[u] = find(ufs[u]);
}
inline void join(int u, int v, int e) {
  u = find(u), v = find(v);
  if (u == v) return;
  ufs[v] = u;
  meme[u] = e;
}
vector<vector<int> > mema;
vector<pair<int, int> > adj[maxn];
vector<int> curm;
inline void dfs(int u) {
  while (adj[u].size()) {
    int v = adj[u].back().first, id = adj[u].back().second;
    adj[u].pop_back();
    dfs(v);
    curm.push_back(id);
  }
}
int main() {
  cin >> n >> s;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    num[++len] = a[i];
  }
  sort(num + 1, num + len + 1);
  len = unique(num + 1, num + len + 1) - num - 1;
  for (int i = 1; i <= n; i++) {
    a[i] = lower_bound(num + 1, num + len + 1, a[i]) - num;
    b[i] = a[i];
    ufs[i] = i;
  }
  sort(b + 1, b + n + 1);
  int cnt = 0;
  for (int i = 1; i <= n; i++) {
    if (a[i] != b[i]) {
      cnt++;
      join(a[i], b[i], i);
    }
  }
  if (cnt > s) {
    cout << "-1" << endl;
    return 0;
  }
  if (cnt == 0) {
    cout << "0" << endl;
    return 0;
  }
  if (s - cnt > 1) {
    for (int i = 1; i <= len; i++) {
      if (ufs[i] == i && meme[i] != 0) {
        curm.push_back(meme[i]);
        if (curm.size() == s - cnt) break;
      }
    }
    if (curm.size() > 1) {
      mema.push_back(curm);
      int lstv = a[curm.back()];
      for (int i = curm.size() - 1; i >= 1; i--) {
        a[curm[i]] = a[curm[i - 1]];
      }
      a[curm[0]] = lstv;
    }
  }
  for (int i = 1; i <= n; i++) {
    if (a[i] != b[i]) {
      adj[a[i]].push_back(make_pair(b[i], i));
    }
  }
  for (int i = 1; i <= len; i++) {
    if (adj[i].size() != 0) {
      curm.clear();
      dfs(i);
      mema.push_back(curm);
    }
  }
  cout << mema.size() << endl;
  for (int i = 0; i < mema.size(); i++) {
    cout << mema[i].size() << endl;
    for (int j = 0; j < mema[i].size(); j++) {
      printf("%d ", mema[i][j]);
    }
    cout << endl;
  }
  return 0;
}