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;
int A[200005], B[200005], Q[200005];
pair<int, int> P[200005];
vector<int> getcycle(int x) {
  int init = x;
  vector<int> ret;
  do {
    ret.push_back(x);
    x = Q[x];
  } while (x != init);
  return ret;
}
int par[200005], pos[200005];
int root(int a) {
  if (par[a] == a) return a;
  return par[a] = root(par[a]);
}
bool join(int x, int y) {
  int a = root(x), b = root(y);
  if (a == b) return false;
  par[a] = b;
  return true;
}
int main() {
  int n, s;
  scanf("%d%d", &n, &s);
  for (int i = 0; i < n; ++i) {
    scanf("%d", &A[i]);
    B[i] = A[i];
  }
  sort(B, B + n);
  int m = 0;
  for (int i = 0; i < n; ++i) {
    if (B[i] != A[i]) {
      P[m] = {A[i], m};
      pos[m] = i + 1;
      ++m;
    }
  }
  sort(P, P + m);
  vector<pair<int, int>> merges;
  for (int i = 0; i < m; ++i) {
    if (i && P[i].first == P[i - 1].first) {
      merges.emplace_back(P[i].second, P[i - 1].second);
    }
    Q[P[i].second] = i;
  }
  vector<bool> used(m);
  iota(par, par + m, 0);
  int numcycles = 0;
  for (int i = 0; i < m; ++i) {
    if (used[i]) continue;
    ++numcycles;
    auto cyc = getcycle(i);
    for (int i = 0; i < cyc.size(); ++i) {
      used[cyc[i]] = true;
      if (i) join(cyc[i - 1], cyc[i]);
    }
  }
  for (auto merge : merges) {
    if (join(merge.first, merge.second)) {
      swap(Q[merge.first], Q[merge.second]);
      --numcycles;
    }
  }
  if (m > s) {
    printf("-1\n");
    return 0;
  }
  int l = s - m;
  vector<vector<int>> ans;
  if (l > 1 && numcycles > 1) {
    fill(used.begin(), used.end(), 0);
    vector<int> firsts;
    for (int i = 0; i < m && firsts.size() < l; ++i) {
      if (used[i]) continue;
      auto cyc = getcycle(i);
      for (int i = 0; i < cyc.size(); ++i) {
        used[cyc[i]] = true;
      }
      firsts.push_back(i);
    }
    ans.push_back(firsts);
    assert(firsts.size() >= 2);
    int qb = Q[firsts.back()];
    for (int i = firsts.size() - 1; i; --i) {
      Q[firsts[i]] = Q[firsts[i - 1]];
    }
    Q[0] = qb;
  }
  fill(used.begin(), used.end(), 0);
  for (int i = 0; i < m; ++i) {
    if (used[i]) continue;
    auto cyc = getcycle(i);
    for (int i = 0; i < cyc.size(); ++i) {
      used[cyc[i]] = true;
    }
    ans.push_back(cyc);
  }
  printf("%d\n", ans.size());
  for (auto &vec : ans) {
    printf("%d\n", vec.size());
    for (int x : vec) {
      printf("%d ", pos[x]);
    }
    printf("\n");
  }
}