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>
int n, s, a[200010], b[200010], nums[200010], cnt, fa[200010];
int find(int i) { return fa[i] == i ? i : fa[i] = find(fa[i]); }
bool tag[200010];
struct edge {
  int to;
  edge* next;
} E[200010], *fir[200010];
std::vector<int> C[200010];
void dfs(int i, int t) {
  while (fir[i]) {
    edge* e = fir[i];
    fir[i] = e->next;
    dfs(e->to, t);
    C[t].push_back(e - E);
  }
}
void mf() {
  for (int i = 0; i < n; i++) fa[i] = i, tag[i] = 0;
  for (int i = 0; i < n; i++)
    if (find(a[i]) != find(b[i]))
      fa[find(a[i])] = find(b[i]), tag[find(b[i])] = 1;
}
void me() {
  for (int i = 0; i < n; i++) fir[i] = 0;
  for (int i = 0; i < n; i++)
    if (a[i] != b[i]) {
      E[i] = (edge){b[i], fir[a[i]]};
      fir[a[i]] = E + i;
    }
}
int rs[200010];
void cyc(int c) {
  if (c < 2) return;
  int v = a[rs[c - 1]];
  for (int i = c; --i;) a[rs[i]] = a[rs[i - 1]];
  a[rs[0]] = v;
}
int main() {
  scanf("%d%d", &n, &s);
  for (int i = 0; i < n; i++)
    scanf("%d", a + i), b[i] = a[i], nums[cnt++] = a[i];
  std::sort(b, b + n);
  std::sort(nums, nums + cnt);
  cnt = std::unique(nums, nums + cnt) - nums;
  for (int i = 0; i < n; i++) {
    a[i] = std::lower_bound(nums, nums + cnt, a[i]) - nums;
    b[i] = std::lower_bound(nums, nums + cnt, b[i]) - nums;
  }
  int t = 0;
  for (int i = 0; i < n; i++)
    if (a[i] != b[i]) t++;
  if (t > s) return puts("-1"), 0;
  mf();
  me();
  int tot = 0;
  for (int i = 0; i < n; i++)
    if (fa[i] == i && tag[i]) rs[tot++] = fir[i] - E;
  if (t + tot > s) tot = s - t;
  cyc(tot);
  mf();
  me();
  for (int i = t = 0; i < n; i++)
    if (fa[i] == i && tag[i]) dfs(i, t++);
  printf("%d\n", t + (tot > 1));
  if (tot > 1) {
    printf("%d\n", tot);
    for (int i = 0; i < tot; i++)
      printf("%d%c", rs[i] + 1, " \n"[i == tot - 1]);
  }
  for (int i = 0; i < t; i++) {
    printf("%d\n", C[i].size());
    for (int j = 0; j < C[i].size(); j++)
      printf("%d%c", C[i][j] + 1, " \n"[j == C[i].size() - 1]);
  }
}