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>
#pragma GCC optimize "-O3"
using namespace std;
const int MAXN = 210000;
int was[MAXN];
int wss[MAXN];
vector<int> st;
vector<vector<int>> vv;
vector<vector<int>> vv2;
int lb[MAXN];
int rb[MAXN];
set<int> ss;
int pl[MAXN], cl[MAXN];
int pos[MAXN];
int n;
void dfs1(int v) {
  was[v] = 1;
  st.push_back(v);
  if (was[pos[v]]) return;
  dfs1(pos[v]);
}
void dfs2(int v, int start) {
  int len = vv[v].size() / 2;
  for (int i = 0; i < len; ++i) ss.erase(vv[v][i]);
  wss[v] = 1;
  for (int i = start; i < start + len; ++i) {
    int x = vv[v][i];
    st.push_back(x);
    auto it = ss.lower_bound(lb[x]);
    if (it != ss.end() && *it < rb[x]) {
      int u = *it;
      dfs2(cl[u], pl[u]);
    }
  }
}
int a[MAXN];
int b[MAXN];
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  int s;
  cin >> n >> s;
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
    b[i] = a[i];
  }
  sort(b, b + n);
  set<pair<int, int>> ss2;
  for (int i = 0; i < n; ++i) {
    if (a[i] == b[i])
      pos[i] = i;
    else {
      ss2.insert({b[i], i});
    }
  }
  for (int i = 0; i < n; ++i) {
    if (a[i] != b[i]) {
      auto it = ss2.lower_bound({a[i], 0});
      pos[i] = it->second;
      ss2.erase(it);
    }
  }
  for (int i = 0; i < n; ++i) {
    lb[i] = lower_bound(b, b + n, a[i]) - b;
    rb[i] = lower_bound(b, b + n, a[i] + 1) - b;
  }
  for (int i = 0; i < n; ++i) {
    if (was[i]) continue;
    if (pos[i] == i) continue;
    st.clear();
    dfs1(i);
    for (int j = 0; j < st.size(); ++j) cl[st[j]] = vv.size(), pl[st[j]] = j;
    vv.push_back(st);
  }
  for (int i = 0; i < vv.size(); ++i) {
    int x = vv[i].size();
    for (int j = 0; j < x; ++j) vv[i].push_back(vv[i][j]);
  }
  for (int i = 0; i < n; ++i) {
    if (a[i] != b[i]) ss.insert(i);
  }
  for (int i = 0; i < vv.size(); ++i) {
    if (wss[i]) continue;
    st.clear();
    dfs2(i, 0);
    vv2.push_back(st);
  }
  int sum = 0;
  for (int i = 0; i < vv2.size(); ++i) sum += vv2[i].size();
  if (sum > s) {
    cout << -1 << "\n";
    return 0;
  }
  int av = min((int)vv2.size(), s - sum);
  if (av <= 1) {
    cout << vv2.size() << "\n";
    for (int i = 0; i < vv2.size(); ++i) {
      cout << vv2[i].size() << "\n";
      for (int j : vv2[i]) cout << j + 1 << " ";
      cout << "\n";
    }
  } else {
    cout << vv2.size() - (av - 1) + 1 << "\n";
    int sum = 0;
    for (int i = 0; i < av; ++i) {
      sum += vv2[i].size();
    }
    cout << sum << "\n";
    for (int i = 0; i < av; ++i) {
      for (int j : vv2[i]) cout << j + 1 << " ";
    }
    cout << "\n";
    for (int i = av; i < vv2.size(); ++i) {
      cout << vv2[i].size() << "\n";
      for (int j : vv2[i]) cout << j + 1 << " ";
      cout << "\n";
    }
    cout << av << "\n";
    for (int i = av - 1; i >= 0; --i) cout << vv2[i][0] + 1 << " ";
    cout << "\n";
  }
  return 0;
}