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 max_n = 200222, inf = 1000111222;
int n, s, cnt, to[max_n];
int a[max_n], b[max_n], pos[max_n];
vector<int> all[max_n];
void compress() {
  pair<int, int> p[max_n];
  for (int i = 0; i < 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[max_n];
int parent[max_n];
void init() {
  for (int i = 1; i <= n; ++i) {
    parent[i] = i;
  }
}
int find_set(int v) {
  if (v == parent[v]) {
    return v;
  }
  return parent[v] = find_set(parent[v]);
}
void union_set(int v1, int v2) {
  v1 = find_set(v1);
  v2 = find_set(v2);
  parent[v1] = v2;
}
void find_all_cycles() {
  cycles.clear();
  bool used[max_n];
  memset(used, 0, sizeof(used));
  for (int i = 0; i < n; ++i) {
    if (used[i] == 0 && to[i] != -1) {
      int pos = i;
      vector<int> cycle;
      while (used[pos] == 0) {
        used[pos] = 1;
        cycle.push_back(pos);
        g[a[pos]].push_back({cycles.size(), pos});
        pos = to[pos];
      }
      cycles.push_back(cycle);
    }
  }
}
int main() {
  scanf("%d%d", &n, &s);
  for (int i = 0; i < n; ++i) {
    scanf("%d", &a[i]);
  }
  compress();
  copy(a, a + n, b);
  sort(b, b + n);
  for (int i = 0; i < n; ++i) {
    if (a[i] != b[i]) {
      all[a[i]].push_back(i);
    }
    to[i] = -1;
  }
  for (int i = 0; i < n; ++i) {
    if (a[i] != b[i]) {
      int p = all[b[i]][pos[b[i]]++];
      to[p] = i;
      ++cnt;
    }
  }
  if (cnt > s) {
    puts("-1");
    return 0;
  }
  find_all_cycles();
  init();
  for (int i = 0; i < n; ++i) {
    if (g[i].size() > 1) {
      const pair<int, int> &first = g[i][0];
      for (const pair<int, int> &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 = cycles.size();
  if (s - cnt) {
    int q = min((int)cycles.size(), s - cnt);
    if (q >= 3) {
      ans += 2 - q;
      printf("%d\n", ans);
      vector<int> v;
      printf("%d\n", q);
      for (int i = 0; i < q; ++i) {
        v.push_back(cycles[i][0]);
        printf("%d ", cycles[i][0] + 1);
      }
      printf("\n");
      int cp = to[v.back()];
      for (int i = v.size() - 1; i > 0; --i) {
        to[v[i]] = to[v[i - 1]];
      }
      to[v[0]] = cp;
      find_all_cycles();
    } else {
      printf("%d\n", ans);
    }
  } else {
    printf("%d\n", ans);
  }
  for (const vector<int> &cycle : cycles) {
    printf("%d\n", cycle.size());
    for (int pos : cycle) {
      printf("%d ", pos + 1);
    }
    printf("\n");
  }
  return 0;
}