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 + 10;
int dsu[N];
int rang[N];
int pred(int a) {
  if (a == dsu[a]) return a;
  return dsu[a] = pred(dsu[a]);
}
void unite(int a, int b) {
  a = pred(a);
  b = pred(b);
  if (a != b) {
    if (rang[a] < rang[b]) {
      swap(a, b);
    }
    dsu[b] = a;
    rang[a] += rang[b];
  }
}
bool connected(int a, int b) { return pred(a) == pred(b); }
bool used[N];
int color;
int t;
int p[N];
vector<int> colors[N];
void dfs(int v) {
  used[v] = 1;
  colors[t].push_back(v);
  if (!used[p[v]]) {
    dfs(p[v]);
  }
}
signed main() {
  int n, s;
  cin >> n >> s;
  for (int i = 0; i < n; i++) {
    dsu[i] = i;
    rang[i] = 1;
  }
  vector<int> a(n);
  vector<pair<int, int> > b;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    b.push_back({a[i], i});
  }
  vector<int> ind(n);
  sort(b.begin(), b.end());
  for (int i = 0; i < n; i++) {
    p[b[i].second] = i;
  }
  for (int i = 0; i < n; i++) {
    if (a[i] == b[i].first && p[i] != i) {
      p[b[i].second] = p[i];
      b[p[i]].second = b[i].second;
      p[i] = i;
      b[i].second = i;
    }
  }
  for (int i = 0; i < n; i++) {
    unite(p[i], i);
  }
  int it = -1;
  for (int i = 0; i < n; i++) {
    if (p[b[i].second] == b[i].second) {
      continue;
    }
    if (it >= 0 && a[it] == a[b[i].second]) {
      int x = it;
      int y = b[i].second;
      if (!connected(x, y)) {
        unite(x, y);
        swap(p[x], p[y]);
      }
    }
    it = b[i].second;
  }
  t = 0;
  for (int i = 0; i < n; i++) {
    if (!used[i] && p[i] != i) {
      dfs(i);
      t++;
    }
  }
  int cnt = 0;
  for (int i = 0; i < t; i++) {
    cnt += colors[i].size();
  }
  if (cnt > s) {
    cout << -1;
    return 0;
  }
  s -= cnt;
  s = min(s, t);
  if (s <= 1) {
    cout << t << "\n";
    for (int i = 0; i < t; i++) {
      cout << colors[i].size() << "\n";
      for (int j = 0; j < colors[i].size(); j++) {
        cout << colors[i][j] + 1 << " ";
      }
      cout << "\n";
    }
    return 0;
  }
  cout << (t - s + 2) << "\n";
  for (int i = 0; i < t - s; i++) {
    cout << colors[i + s].size() << "\n";
    for (int j = 0; j < colors[i + s].size(); j++) {
      cout << colors[i + s][j] + 1 << " ";
    }
    cout << "\n";
    cnt -= colors[i + s].size();
  }
  cout << cnt << "\n";
  for (int i = 0; i < s; i++) {
    for (int j = 0; j < colors[i].size(); j++) {
      cout << colors[i][j] + 1 << " ";
    }
  }
  cout << "\n";
  cout << s << "\n";
  for (int i = s - 1; i >= 0; i--) {
    cout << colors[i][0] + 1 << " ";
  }
  return 0;
}