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 Maxn = 200005;
int n, s, ct, tmp_n, ans_ct, a[Maxn], b[Maxn], fa[Maxn], pos[Maxn], ord[Maxn],
    bel[Maxn], Pos[Maxn];
vector<int> spec, Ve[Maxn];
bool vis[Maxn];
void dfs(int u) {
  if (vis[u]) return;
  bel[u] = ct, vis[u] = true, dfs(ord[u]);
}
void dfs2(int u) {
  if (vis[u]) return;
  Ve[ans_ct].push_back(u), vis[u] = true, dfs2(pos[u]);
}
int get_fa(int x) { return x == fa[x] ? x : fa[x] = get_fa(fa[x]); }
int main() {
  scanf("%d%d", &n, &s);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]), b[i] = a[i];
  sort(b + 1, b + 1 + n);
  for (int i = 1; i <= n; i++)
    if (a[i] != b[i]) a[++tmp_n] = a[i], Pos[tmp_n] = i;
  n = tmp_n;
  if (s < n) {
    puts("-1");
    return 0;
  }
  s -= n;
  for (int i = 1; i <= n; i++) ord[i] = i;
  sort(ord + 1, ord + 1 + n, [](int x, int y) { return a[x] < a[y]; });
  for (int i = 1; i <= n; i++) pos[ord[i]] = i;
  a[n + 1] = -1;
  for (int i = 1; i <= n; i++)
    if (!vis[i]) ct++, fa[ct] = ct, dfs(i);
  int las = 1;
  for (int i = 2; i <= n + 1; i++)
    if (a[ord[i]] != a[ord[i - 1]]) {
      for (int j = las; j < i; j++)
        if (get_fa(bel[ord[las]]) != get_fa(bel[ord[j]]))
          fa[get_fa(bel[ord[j]])] = get_fa(bel[ord[las]]),
          swap(ord[j], ord[las]);
      las = i;
    }
  memset(vis, 0, sizeof(bool[n + 1]));
  ct = 0;
  for (int i = 1; i <= n; i++) pos[ord[i]] = i;
  for (int i = 1; i <= n; i++)
    if (!vis[i]) {
      ct++;
      if (ct == 1 || ct > s) ++ans_ct;
      if (ct <= s) spec.push_back(i);
      dfs2(i);
    }
  printf("%d\n", ans_ct + (spec.size() > 1));
  if (ans_ct) {
    printf("%d\n", (int)Ve[1].size());
    for (vector<int>::iterator it = Ve[1].begin(); it != Ve[1].end(); it++)
      printf("%d ", Pos[*it]);
    puts("");
  }
  if (spec.size() > 1) {
    printf("%d\n", (int)spec.size());
    for (vector<int>::reverse_iterator it = spec.rbegin(); it != spec.rend();
         it++)
      printf("%d ", Pos[*it]);
    puts("");
  }
  for (int i = 2; i <= ans_ct; i++) {
    printf("%d\n", (int)Ve[i].size());
    for (vector<int>::iterator it = Ve[i].begin(); it != Ve[i].end(); it++)
      printf("%d ", Pos[*it]);
    puts("");
  }
  return 0;
}