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;
template <class T>
bool uin(T &a, T b) {
  return a > b ? (a = b, true) : false;
}
template <class T>
bool uax(T &a, T b) {
  return a < b ? (a = b, true) : false;
}
const int N = 2e5 + 5;
int n, s, cnt, to[N], a[N], b[N], pos[N];
vector<int> all[N];
pair<int, int> p[N];
void compress() {
  for (int i = 0; i < (int)(n); ++i) p[i] = {a[i], i};
  sort(p, p + n);
  int num = 0;
  for (int i = 0; i < n;) {
    int start = i;
    while (i < n && p[i].first == p[start].first) {
      a[p[i].second] = num;
      i++;
    }
    num++;
  }
}
vector<vector<int> > cycles;
vector<pair<int, int> > g[N];
int par[N];
void init() {
  for (int i = 1; i <= (int)(n); ++i) par[i] = i;
}
int find_set(int v) {
  if (v == par[v]) return v;
  return par[v] = find_set(par[v]);
}
void union_set(int a, int b) {
  a = find_set(a);
  b = find_set(b);
  par[a] = b;
}
bool used[N];
void find_all_cycles() {
  cycles.clear();
  memset(used, 0, sizeof(used));
  for (int i = 0; i < (int)(n); ++i) {
    if (!used[i] && to[i] != -1) {
      int pos = i;
      vector<int> cycle;
      while (used[pos] == 0) {
        used[pos] = 1;
        cycle.push_back(pos);
        g[a[pos]].emplace_back((int)(cycles).size(), pos);
        pos = to[pos];
      }
      cycles.push_back(cycle);
    }
  }
}
int main(int argc, char **argv) {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> s;
  for (int i = 0; i < (int)(n); ++i) cin >> a[i];
  compress();
  copy(a, a + n, b);
  sort(b, b + n);
  for (int i = 0; i < (int)(n); ++i) {
    if (a[i] != b[i]) all[a[i]].push_back(i);
    to[i] = -1;
  }
  for (int i = 0; i < (int)(n); ++i) {
    if (a[i] != b[i]) {
      int p = all[b[i]][pos[b[i]]++];
      to[p] = i;
      cnt++;
    }
  }
  if (cnt > s) {
    cout << -1 << '\n';
    return 0;
  }
  find_all_cycles();
  init();
  for (int i = 0; i < (int)(n); ++i) {
    if ((int)(g[i]).size() > 1) {
      pair<int, int> first = g[i][0];
      for (auto &(p) : (g[i])) {
        if (find_set(first.first) != find_set(p.first)) {
          union_set(first.first, p.first);
          swap(to[first.second], to[p.second]);
        }
      }
    }
  }
  find_all_cycles();
  int ans = (int)(cycles).size();
  if (s > cnt) {
    int q = min((int)(cycles).size(), s - cnt);
    if (q >= 3) {
      ans += 2 - q;
      cout << ans << '\n';
      vector<int> v;
      cout << q << '\n';
      for (int i = 0; i < (int)(q); ++i) {
        v.push_back(cycles[i][0]);
        cout << cycles[i][0] + 1 << " \n"[i == q - 1];
      }
      int cp = to[v.back()];
      for (int i = (int)((int)(v).size() - 1); i >= (int)(1); --i)
        to[v[i]] = to[v[i - 1]];
      to[v[0]] = cp;
      find_all_cycles();
    } else {
      cout << ans << '\n';
    }
  } else
    cout << ans << '\n';
  for (auto &(p) : (cycles)) {
    cout << (int)(p).size() << '\n';
    for (auto &(i) : (p)) cout << i + 1 << ' ';
    cout << '\n';
  }
}