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 = 200600;
int n, S;
int a[N];
int b[N];
int xs[N];
int k;
vector<int> g[N];
int m;
vector<int> cycles[N];
int p[N];
bool used[N];
void dfs(int v) {
  while (!g[v].empty()) {
    int id = g[v].back();
    g[v].pop_back();
    dfs(a[id]);
    cycles[m].push_back(id);
  }
}
int main() {
  scanf("%d%d", &n, &S);
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    b[i] = a[i];
  }
  sort(b, b + n);
  for (int i = 0; i < n; i++) xs[k++] = b[i];
  k = unique(xs, xs + k) - xs;
  for (int i = 0; i < n; i++) {
    a[i] = lower_bound(xs, xs + k, a[i]) - xs;
    b[i] = lower_bound(xs, xs + k, b[i]) - xs;
  }
  for (int i = 0; i < n; i++) {
    if (a[i] == b[i]) continue;
    g[b[i]].push_back(i);
  }
  for (int i = 0; i < k; i++) {
    dfs(i);
    if (!cycles[m].empty()) m++;
  }
  for (int i = 0; i < m; i++) S -= (int)cycles[i].size();
  if (S < 0) {
    printf("-1\n");
    return 0;
  }
  for (int i = 0; i < n; i++) p[i] = i;
  for (int id = 0; id < m; id++) {
    for (int i = 0; i < (int)cycles[id].size(); i++) {
      int v = cycles[id][i], u = cycles[id][(i + 1) % (int)cycles[id].size()];
      p[u] = v;
    }
  }
  S = min(S, m);
  if (S > 1) {
    printf("%d\n", 2 + m - S);
    printf("%d\n", S);
    for (int i = 0; i < S; i++) printf("%d ", cycles[i][0] + 1);
    printf("\n");
    for (int i = S - 1; i > 0; i--) swap(p[cycles[i][0]], p[cycles[i - 1][0]]);
  } else {
    printf("%d\n", m);
  }
  for (int i = 0; i < n; i++) {
    if (used[i]) continue;
    if (p[i] == i) {
      used[i] = 1;
      continue;
    }
    vector<int> cur;
    int x = i;
    while (!used[x]) {
      cur.push_back(x);
      used[x] = 1;
      x = p[x];
    }
    printf("%d\n", (int)cur.size());
    for (int z : cur) printf("%d ", z + 1);
    printf("\n");
  }
  return 0;
}