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 n, K;
map<int, int> Map;
int A[201000], B[201000], w[201000];
vector<int> GG[201000];
int GPV[201000];
int tA[201000];
vector<int> G;
int UF[201000], sss;
int Find(int a) {
  if (a == UF[a]) return a;
  return UF[a] = Find(UF[a]);
}
void Merge(int a, int b) {
  a = Find(a), b = Find(b);
  if (a != b) UF[a] = b;
}
int v[201000];
vector<vector<int> > RR;
void Pro(vector<int> &T) {
  RR.push_back(T);
  for (int i = T.size() - 1; i >= 0; i--) {
    int ni = i - 1;
    if (ni < 0) ni = T.size() - 1;
    tA[T[ni]] = A[T[i]];
  }
  for (auto &t : T) A[t] = tA[t];
}
void Go(int a) {
  vector<int> T;
  int t = a;
  while (1) {
    v[t] = 1;
    T.push_back(t);
    t = w[t];
    if (t == a) break;
  }
  Pro(T);
  return;
}
bool Do(int ck) {
  int i, c = 0;
  for (i = 1; i <= n; i++) {
    B[i] = A[i];
    GG[i].clear();
    UF[i] = i;
    GPV[i] = 0;
    v[i] = 0;
  }
  sort(B + 1, B + n + 1);
  Map.clear();
  for (i = 1; i <= n; i++) {
    if (A[i] != B[i]) c++;
  }
  if (c > K) {
    puts("-1");
    return false;
  }
  for (i = 1; i <= n; i++) {
    if (A[i] != B[i]) {
      if (!Map.count(A[i])) {
        Map[A[i]] = i;
        GG[i].push_back(i);
      } else {
        GG[Map[A[i]]].push_back(i);
      }
    }
  }
  for (i = 1; i <= n; i++) {
    if (A[i] != B[i]) {
      G.push_back(i);
      int x = Map[B[i]];
      w[i] = GG[x][GPV[x]];
      GPV[x]++;
    }
  }
  for (auto &x : G) {
    Merge(x, w[x]);
  }
  for (i = 0; i + 1 < G.size(); i++) {
    int x = G[i], y = G[i + 1];
    if (B[x] == B[y] && Find(x) != Find(y)) {
      swap(w[x], w[y]);
      Merge(x, y);
    }
  }
  int res = 0;
  vector<int> TP;
  for (auto &x : G) {
    if (!v[Find(x)]) {
      TP.push_back(x);
      res++;
      v[Find(x)] = 1;
    }
  }
  if (!ck) {
    int cur = G.size();
    int rr = min(res, max(K - cur, 0));
    if (rr >= 3) {
      res = res - rr + 2;
      vector<int> TP2;
      for (int i = 0; i < rr; i++) {
        TP2.push_back(TP[i]);
      }
      Pro(TP2);
      return Do(1);
    }
  }
  for (i = 1; i <= n; i++) v[i] = 0;
  for (auto &x : G) {
    if (!v[x]) {
      Go(x);
    }
  }
  return true;
}
int main() {
  int i;
  scanf("%d%d", &n, &K);
  for (i = 1; i <= n; i++) {
    scanf("%d", &A[i]);
  }
  if (Do(0)) {
    printf("%d\n", RR.size());
    for (auto &t : RR) {
      printf("%d\n", t.size());
      for (int i = t.size() - 1; i >= 0; i--) printf("%d ", t[i]);
      puts("");
    }
  }
}