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 BidirIt>
BidirIt prev(BidirIt it,
             typename iterator_traits<BidirIt>::difference_type n = 1) {
  advance(it, -n);
  return it;
}
template <class ForwardIt>
ForwardIt next(ForwardIt it,
               typename iterator_traits<ForwardIt>::difference_type n = 1) {
  advance(it, n);
  return it;
}
const double EPS = 1e-9;
const double PI = 3.141592653589793238462;
template <typename T>
inline T sq(T a) {
  return a * a;
}
const int MAXN = 4e5 + 5;
int ar[MAXN], sor[MAXN];
map<int, int> dummy;
bool visit[MAXN], proc[MAXN];
int nxt[MAXN];
vector<int> gr[MAXN];
vector<int> cur, tour[MAXN];
vector<vector<int> > cycles;
void addEdge(int u, int v) { gr[u].push_back(v); }
void dfs(int u) {
  visit[u] = true;
  while (nxt[u] < (int)gr[u].size()) {
    int v = gr[u][nxt[u]];
    nxt[u]++;
    dfs(v);
    cur.push_back(u);
  }
}
void getcycle(int u, vector<int> &vec) {
  proc[u] = true;
  for (auto it : tour[u]) {
    vec.push_back(it);
    if (!proc[it]) getcycle(it, vec);
  }
}
int main() {
  int n, s;
  scanf("%d %d", &n, &s);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &ar[i]);
    sor[i] = ar[i];
  }
  sort(sor + 1, sor + n + 1);
  for (int i = 1; i <= n; i++) {
    if (ar[i] != sor[i]) dummy[ar[i]] = 0;
  }
  int cnt = 0;
  for (auto &it : dummy) it.second = n + (++cnt);
  for (int i = 1; i <= n; i++) {
    if (ar[i] == sor[i]) continue;
    addEdge(dummy[sor[i]], i);
    addEdge(i, dummy[ar[i]]);
  }
  for (int i = 1; i <= n + cnt; i++) {
    if ((i > n || ar[i] != sor[i]) && !visit[i]) {
      cur.clear();
      dfs(i);
      reverse((cur).begin(), (cur).end());
      vector<int> res;
      for (auto it : cur)
        if (it <= n) res.push_back(it);
      cycles.push_back(res);
      s -= (int)res.size();
    }
  }
  if (s < 0) {
    puts("-1");
    return 0;
  }
  int pos = 0;
  if (s > 1 && cycles.size() > 1) {
    printf("%d\n", max(2, (int)cycles.size() - s + 2));
    int sum = 0;
    while (s > 0 && pos < (int)cycles.size()) {
      s--;
      sum += (int)cycles[pos].size();
      pos++;
    }
    printf("%d\n", sum);
    vector<int> vec;
    for (int i = 0; i < pos; i++) {
      for (auto it : cycles[i]) printf("%d ", it);
      vec.push_back(cycles[i][0]);
    }
    puts("");
    reverse((vec).begin(), (vec).end());
    printf("%d\n", (int)vec.size());
    for (auto it : vec) printf("%d ", it);
    puts("");
  } else {
    printf("%d\n", (int)cycles.size());
  }
  for (int i = pos; i < (int)cycles.size(); i++) {
    printf("%d\n", (int)cycles[i].size());
    for (auto it : cycles[i]) printf("%d ", it);
    puts("");
  }
  return 0;
}