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;
template <typename T>
inline void ckmax(T& x, T y) {
  x = (y > x ? y : x);
}
template <typename T>
inline void ckmin(T& x, T y) {
  x = (y < x ? y : x);
}
const int MAXN = 2e5;
int n, lim, a[MAXN + 5], goal[MAXN + 5];
int vals[MAXN + 5], cnt_val;
int need_oper_pos;
vector<vector<int> > ans;
int fa[MAXN + 5], sz[MAXN + 5];
int last_edge[MAXN + 5];
int get_fa(int u) { return (u == fa[u]) ? u : (fa[u] = get_fa(fa[u])); }
void unite(int u, int v, int e) {
  int fu = get_fa(u);
  int fv = get_fa(v);
  if (fu != fv) {
    if (sz[fu] > sz[fv]) swap(fu, fv);
    fa[fu] = fv;
    sz[fv] += sz[fu];
  }
  last_edge[fu] = e;
}
struct EDGE {
  int nxt, to, eid;
} edge[MAXN + 5];
int head[MAXN + 5], tot;
inline void add_edge(int u, int v, int eid) {
  edge[++tot].nxt = head[u];
  edge[tot].to = v;
  edge[tot].eid = eid;
  head[u] = tot;
}
vector<int> pos;
void dfs(int u) {
  for (int& i = head[u]; i;) {
    int v = edge[i].to;
    int eid = edge[i].eid;
    i = edge[i].nxt;
    dfs(v);
    pos.push_back(eid);
  }
}
int main() {
  cin >> n >> lim;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
    vals[++cnt_val] = a[i];
  }
  sort(vals + 1, vals + cnt_val + 1);
  for (int i = 1; i <= n; ++i) goal[i] = vals[i];
  cnt_val = unique(vals + 1, vals + cnt_val + 1) - (vals + 1);
  for (int i = 1; i <= cnt_val; ++i) {
    fa[i] = i;
    sz[i] = 1;
  }
  for (int i = 1; i <= n; ++i) {
    a[i] = lower_bound(vals + 1, vals + cnt_val + 1, a[i]) - vals;
    goal[i] = lower_bound(vals + 1, vals + cnt_val + 1, goal[i]) - vals;
    if (a[i] != goal[i]) {
      unite(a[i], goal[i], i);
      need_oper_pos++;
    }
  }
  if (need_oper_pos > lim) {
    cout << -1 << endl;
    return 0;
  }
  int rest = lim - need_oper_pos;
  if (rest >= 2) {
    for (int i = 1; i <= cnt_val; ++i) {
      if (get_fa(i) == i && last_edge[i] != 0) {
        pos.push_back(last_edge[i]);
        if (((int)(pos).size()) == rest) break;
      }
    }
    if (((int)(pos).size()) >= 2) {
      ans.push_back(pos);
      int last_val = a[pos.back()];
      for (int i = ((int)(pos).size()) - 1; i >= 1; --i) {
        a[pos[i]] = a[pos[i - 1]];
      }
      a[pos[0]] = last_val;
    }
  }
  for (int i = 1; i <= n; ++i) {
    if (a[i] != goal[i]) {
      add_edge(a[i], goal[i], i);
    }
  }
  for (int i = 1; i <= n; ++i) {
    if (head[i]) {
      vector<int>().swap(pos);
      dfs(i);
      ans.push_back(pos);
    }
  }
  cout << ((int)(ans).size()) << endl;
  for (int i = 0; i < ((int)(ans).size()); ++i) {
    cout << ((int)(ans[i]).size()) << endl;
    for (int j = 0; j < ((int)(ans[i]).size()); ++j) {
      cout << ans[i][j] << " \n"[j == ((int)(ans[i]).size()) - 1];
    }
  }
  return 0;
}