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;
namespace FGF {
int n, m;
const int N = 2e5 + 5;
int a[N], b[N], p[N], vis[N], cnt, head[N], fa[N], ins[N], ok[N];
vector<int> V, tmp;
struct edg {
  int to, nxt, w;
} e[N];
void add(int u, int v, int w) {
  cnt++;
  e[cnt].to = v;
  e[cnt].nxt = head[u];
  head[u] = cnt;
  e[cnt].w = w;
}
int find(int x) { return fa[x] == x ? x : fa[x] = find(fa[x]); }
void merge(int u, int v) {
  if (find(u) != find(v)) fa[find(v)] = u;
}
vector<vector<int> > ans;
void dfs(int u) {
  ins[u] = 1;
  for (int w, &i = head[u]; i; i = e[i].nxt)
    if (!vis[i]) vis[i] = 1, w = e[i].w, dfs(e[i].to), tmp.push_back(w);
}
void work() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]), b[i] = a[i];
  sort(b + 1, b + n + 1);
  int tot = unique(b + 1, b + n + 1) - b - 1;
  for (int i = 1; i <= n; i++) a[i] = lower_bound(b + 1, b + tot + 1, a[i]) - b;
  for (int i = 1; i <= n; i++) p[i] = a[i], fa[i] = i;
  sort(p + 1, p + n + 1);
  for (int i = 1; i <= n; i++)
    if (a[i] != p[i]) cnt++, merge(a[i], p[i]);
  if (cnt > m) {
    puts("-1");
    return;
  }
  for (int i = 1; i <= n; i++)
    if (find(a[i]) == a[i] && a[i] != p[i] && !ok[a[i]])
      V.push_back(i), ok[a[i]] = 1;
  if (m - cnt > 1 && V.size() > 1) {
    tmp.push_back(V[0]);
    for (int i = 1; i < min((int)V.size(), m - cnt); i++)
      swap(a[V[i]], a[V[i - 1]]), tmp.push_back(V[i]);
    reverse(tmp.begin(), tmp.end());
    ans.push_back(tmp);
    tmp.clear();
  }
  cnt = 0;
  for (int i = 1; i <= n; i++)
    if (a[i] != p[i]) add(a[i], p[i], i);
  for (int i = 1; i <= tot; i++)
    if (!ins[i] && head[i]) tmp.clear(), dfs(i), ans.push_back(tmp);
  printf("%d\n", (int)ans.size());
  for (auto g : ans) {
    printf("%d\n", (int)g.size());
    for (auto x : g) printf("%d ", x);
    puts("");
  }
}
}  // namespace FGF
int main() {
  FGF::work();
  return 0;
}