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 + 20;
int a[maxn], q[maxn], num[maxn], par[maxn];
bool visited[maxn];
vector<int> ind[maxn], cycle[maxn];
int f(int v) { return (par[v] == -1) ? v : par[v] = f(par[v]); }
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  memset(par, -1, sizeof par);
  int n, s;
  cin >> n >> s;
  vector<int> tmp;
  for (int i = 0; i < n; i++) cin >> a[i], tmp.push_back(a[i]);
  sort(tmp.begin(), tmp.end());
  tmp.resize(unique(tmp.begin(), tmp.end()) - tmp.begin());
  for (int i = 0; i < n; i++)
    a[i] = lower_bound(tmp.begin(), tmp.end(), a[i]) - tmp.begin();
  for (int i = 0; i < n; i++) ind[a[i]].push_back(i);
  int t = 0;
  for (int i = 0; i < (int)tmp.size(); i++) {
    int sz = ind[i].size();
    vector<int> tmp2 = ind[i];
    for (auto &x : ind[i])
      if (t <= x && x < t + sz) {
        a[x] = x;
        visited[x] = 1;
        x = 1e9;
      }
    sort(ind[i].begin(), ind[i].end());
    while (!ind[i].empty() && ind[i].back() > n) ind[i].pop_back();
    tmp2 = ind[i];
    for (int j = t; j < t + sz; j++)
      if (!visited[j]) a[tmp2.back()] = j, tmp2.pop_back();
    t += sz;
  }
  t = 0;
  memset(par, -1, sizeof par);
  for (int i = 0; i < n; i++)
    if (!visited[i]) {
      while (!visited[i]) {
        visited[i] = 1;
        cycle[t].push_back(i);
        i = a[i];
      }
      for (int j = 1; j < (int)cycle[t].size(); j++)
        par[cycle[t][j]] = cycle[t][0];
      t++;
    }
  for (int i = 0; i < (int)tmp.size(); i++) {
    if (ind[i].empty()) continue;
    int marja = ind[i].back();
    ind[i].pop_back();
    for (auto shit : ind[i]) {
      if (f(shit) == f(marja)) continue;
      par[f(shit)] = f(marja);
      swap(a[shit], a[marja]);
    }
  }
  memset(visited, 0, sizeof visited);
  for (int i = 0; i < t; i++) cycle[i].clear();
  t = 0;
  int T = 0;
  for (int i = 0; i < n; i++)
    if (!visited[i]) {
      int sz = 0;
      while (!visited[i]) {
        sz++;
        cycle[t].push_back(i);
        visited[i] = 1;
        i = a[i];
      }
      if (sz != 1)
        t++;
      else
        cycle[t].clear();
    }
  if (!t) return cout << 0 << endl, 0;
  if (t == 1) {
    if (s < (int)cycle[0].size()) return cout << -1 << endl, 0;
    cout << 1 << endl;
    cout << cycle[0].size() << endl;
    for (auto x : cycle[0]) cout << x + 1 << " ";
    cout << endl;
    return 0;
  }
  for (int i = 0; i < t; i++) T += (int)cycle[i].size();
  if (s >= T + t) {
    cout << 2 << endl;
    cout << T << endl;
    for (int i = 0; i < t; i++)
      for (auto x : cycle[i]) cout << x + 1 << " ";
    cout << endl;
    cout << t << endl;
    for (int i = t - 1; i >= 0; i--) cout << cycle[i][0] + 1 << " ";
    cout << endl;
    return 0;
  }
  if (s < T) return cout << -1 << endl, 0;
  s -= T;
  if (s < 2) {
    cout << t << endl;
    for (int i = 0; i < t; i++) {
      cout << cycle[i].size() << endl;
      for (auto x : cycle[i]) cout << x + 1 << " ";
      cout << endl;
    }
    return 0;
  }
  cout << 2 + (t - s) << endl;
  for (int i = s; i < t; i++) {
    cout << cycle[i].size() << endl;
    for (auto x : cycle[i]) cout << x + 1 << " ";
    cout << endl;
  }
  t = s;
  T = 0;
  for (int i = 0; i < t; i++) T += (int)cycle[i].size();
  cout << T << endl;
  for (int i = 0; i < t; i++)
    for (auto x : cycle[i]) cout << x + 1 << " ";
  cout << endl;
  cout << t << endl;
  for (int i = t - 1; i >= 0; i--) cout << cycle[i][0] + 1 << " ";
  cout << endl;
}