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 = 1e6 + 7;
const int MX = 1e9 + 7;
const long long int INF = 1e18 + 9LL;
int n, s;
int in[N];
int last[N];
int sorted[N];
set<int> place[N];
set<int> value[N];
int cnt;
map<int, int> M;
int dir[N];
bool vis[N];
vector<int> cur;
void solve(int c) {
  if (place[c].size() == 0) return;
  int v1 = *place[c].begin(), v2 = *value[c].begin();
  place[c].erase(v1);
  value[c].erase(v2);
  int first = v1;
  dir[v1] = v2;
  if (last[in[first]] > 0) swap(dir[first], dir[last[in[first]]]);
  last[in[first]] = first;
  vector<int> cycle;
  cycle.push_back(v1);
  while (v2 != first) {
    v1 = v2;
    last[in[v1]] = v1;
    cycle.push_back(v1);
    c = in[v1];
    place[c].erase(v1);
    v2 = *value[c].begin();
    value[c].erase(v2);
    dir[v1] = v2;
  }
  for (int v : cycle)
    if (place[in[v]].size()) solve(in[v]);
}
void dfs(int u) {
  vis[u] = true;
  cur.push_back(u);
  if (!vis[dir[u]]) dfs(dir[u]);
}
int main() {
  scanf("%d %d", &n, &s);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &in[i]);
    sorted[i] = in[i];
  }
  int inc = 0;
  sort(sorted + 1, sorted + n + 1);
  for (int i = 1; i <= n; ++i) {
    if (in[i] != sorted[i]) ++inc;
    if (!M.count(sorted[i])) M[sorted[i]] = ++cnt;
  }
  if (inc > s) {
    puts("-1");
    return 0;
  }
  for (int i = 1; i <= n; ++i) {
    in[i] = M[in[i]];
    sorted[i] = M[sorted[i]];
  }
  for (int i = 1; i <= n; ++i)
    if (in[i] != sorted[i]) {
      place[in[i]].insert(i);
      value[sorted[i]].insert(i);
    }
  vector<vector<int> > ans;
  for (int i = 1; i <= n; ++i) {
    if (place[i].size() == 0) continue;
    solve(i);
  }
  for (int i = 1; i <= n; ++i) vis[i] = in[i] == sorted[i];
  for (int i = 1; i <= n; ++i)
    if (!vis[i]) {
      cur.clear();
      dfs(i);
      ans.push_back(cur);
    }
  s -= inc;
  if (s >= 3 && ans.size() >= 3) {
    s = min(s, (int)ans.size());
    vector<int> help, help2;
    while (s--) {
      for (auto v : ans.back()) help.push_back(v);
      help2.push_back(ans.back()[0]);
      ans.pop_back();
    }
    reverse(help2.begin(), help2.end());
    ans.push_back(help);
    ans.push_back(help2);
  }
  printf("%d\n", (int)ans.size());
  for (auto V : ans) {
    printf("%d\n", (int)V.size());
    for (auto v : V) printf("%d ", v);
    puts("");
  }
  return 0;
}