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;
long long read() {
  long long x = 0, F = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') F = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = x * 10 + c - '0';
    c = getchar();
  }
  return x * F;
}
int n, cnt, s, F, tmp;
int a[500000 + 5], b[500000 + 5], val[500000 + 5];
int head[500000 + 5], ecnt;
struct edge {
  int to, nxt, vis;
} e[500000 + 5];
vector<int> P[500000 + 5];
void link(int u, int v) { e[++ecnt] = (edge){v, head[u]}, head[u] = ecnt; }
void dfs(int x) {
  for (int &i = head[x]; i;) {
    int v = e[i].to;
    i = e[i].nxt;
    dfs(a[v]);
    P[cnt].push_back(v);
  }
}
void Print(int id) {
  printf("%d\n", P[id].size());
  for (int i = P[id].size() - 1; i >= 0; i--) printf("%d ", P[id][i]);
  puts("");
}
int main() {
  n = read(), s = read();
  for (int i = 1; i <= n; i++) val[i] = a[i] = read();
  sort(val + 1, val + n + 1);
  int pn = unique(val + 1, val + n + 1) - val - 1;
  for (int i = 1; i <= n; i++)
    b[i] = a[i] = lower_bound(val + 1, val + pn + 1, a[i]) - val;
  sort(b + 1, b + n + 1);
  for (int i = 1; i <= n; i++)
    if (b[i] != a[i]) link(b[i], i), s--, F = 1;
  if (!F) {
    puts("0");
    return 0;
  } else if (s < 0) {
    puts("-1");
    return 0;
  }
  for (int i = 1; i <= pn; i++)
    if (head[i]) cnt++, dfs(i);
  if (cnt == 1) {
    puts("1");
    Print(1);
    return 0;
  }
  s = min(s, cnt);
  if (s > 1)
    printf("%d\n", cnt - s + 2);
  else
    printf("%d\n", cnt);
  for (int i = cnt; i > s; i--) Print(i);
  if (s) {
    int tot = 0;
    for (int i = 1; i <= s; i++) tot += P[i].size();
    printf("%d\n", tot);
    for (int i = s; i >= 1; i--)
      for (int j = P[i].size() - 1; j >= 0; j--) printf("%d ", P[i][j]);
    puts("");
    if (s > 1) {
      printf("%d\n", s);
      for (int i = 1; i <= s; i++) printf("%d ", P[i][P[i].size() - 1]);
    }
  }
}