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 MN = 200010;
int fa[MN];
void init() {
  for (int i = 0; i < MN; i++) fa[i] = i;
}
int find(int u) {
  if (fa[u] == u)
    return u;
  else
    return fa[u] = find(fa[u]);
}
void mrg(int u, int v) {
  u = find(u);
  v = find(v);
  if (u == v) return;
  fa[v] = u;
}
int N, S;
int A[MN], B[MN], P[MN], vis[MN];
int Xn;
vector<int> X;
unordered_map<int, int> dx;
vector<int> Z[MN];
set<int> V[MN];
vector<vector<int> > sol;
int main() {
  scanf("%d %d", &N, &S);
  for (int i = 0; i < N; i++) {
    scanf("%d", &A[i]);
    X.push_back(A[i]);
  }
  sort(X.begin(), X.end());
  X.resize(unique(X.begin(), X.end()) - X.begin());
  Xn = X.size();
  for (int i = 0; i < Xn; i++) dx[X[i]] = i;
  for (int i = 0; i < N; i++) A[i] = dx[A[i]];
  for (int i = 0; i < N; i++) {
    B[i] = A[i];
    V[A[i]].insert(i);
  }
  sort(B, B + N);
  int cnt = 0;
  for (int i = 0; i < N; i++) {
    if (A[i] == B[i]) {
      P[i] = i;
      V[A[i]].erase(i);
    } else {
      cnt++;
      Z[A[i]].push_back(i);
    }
  }
  if (cnt > S) {
    printf("-1");
    return 0;
  }
  for (int i = 0; i < N; i++) {
    if (A[i] != B[i]) {
      P[*V[B[i]].begin()] = i;
      V[B[i]].erase(V[B[i]].begin());
    }
  }
  init();
  for (int i = 0; i < N; i++) {
    mrg(i, P[i]);
  }
  for (int i = 0; i < Xn; i++) {
    for (int j = 0; j < (int)Z[i].size() - 1; j++) {
      int u = Z[i][j];
      int v = Z[i][j + 1];
      if (find(u) != find(v)) {
        swap(P[u], P[v]);
        mrg(u, v);
      }
    }
  }
  vector<int> tmp, pre;
  for (int i = 0; i < N; i++)
    if (P[i] != i && fa[i] == i) {
      tmp.push_back(i);
    }
  int n = min(S - cnt, (int)tmp.size());
  if (n > 1) {
    for (int i = 0; i < n; i++) {
      pre.push_back(P[tmp[i]]);
    }
    sol.push_back(vector<int>());
    for (int i = 0; i < n; i++) {
      sol.back().push_back(tmp[i]);
      P[tmp[i]] = pre[(i + n - 1) % n];
    }
  }
  for (int i = 0; i < N; i++) {
    if (vis[i]) continue;
    if (P[i] == i) continue;
    sol.push_back(vector<int>());
    int u = i;
    while (!vis[u]) {
      vis[u] = 1;
      sol.back().push_back(u);
      u = P[u];
    }
  }
  printf("%d\n", sol.size());
  for (int i = 0; i < sol.size(); i++) {
    printf("%d\n", sol[i].size());
    for (int j = 0; j < sol[i].size(); j++) {
      printf("%d ", sol[i][j] + 1);
    }
    printf("\n");
  }
}