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;
int n, S, a[200010], fa[200010];
inline int getf(int x) { return x == fa[x] ? x : fa[x] = getf(fa[x]); }
inline void merge(int x, int y) {
  x = getf(x);
  y = getf(y);
  if (x != y) fa[x] = y;
}
pair<int, int> p[200010];
class node {
 public:
  int vl, pos, id;
  bool operator<(const node &t) const { return vl < t.vl; }
};
int rk[200010], rr[200010];
bool vis[200010];
int main() {
  scanf("%d%d", &n, &S);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    p[i] = make_pair(a[i], i);
  }
  sort(p + 1, p + n + 1);
  iota(fa + 1, fa + n + 1, 1);
  int cnt = 0;
  for (int i = 1; i <= n; i++)
    if (a[i] != p[i].first) cnt++;
  if (cnt > S) {
    puts("-1");
    return 0;
  }
  vector<node> v;
  v.push_back((node){0, 0, 0});
  for (int i = 1; i <= n; i++) {
    if (a[i] != p[i].first) {
      v.push_back((node){a[i], 0, i});
      v.back().pos = v.size() - 1;
    }
  }
  sort(v.begin() + 1, v.end());
  int s = v.size() - 1;
  for (int i = 1; i <= s; i++) rk[v[i].pos] = i, rr[v[i].pos] = v[i].id;
  for (int i = 1; i <= s; i++) merge(i, rk[i]);
  for (int i = 2; i <= s; i++) {
    if (v[i].vl == v[i - 1].vl && getf(i) != getf(i - 1)) {
      swap(rk[v[i].pos], rk[v[i - 1].pos]);
      merge(i - 1, i);
    }
  }
  vector<vector<int> > cycle;
  for (int i = 1; i <= s; i++) {
    if (!vis[i]) {
      cycle.push_back(vector<int>({rr[i]}));
      vis[i] = 1;
      for (int j = rk[i]; j != i; j = rk[j]) {
        cycle.back().push_back(rr[j]);
        vis[j] = 1;
      }
    }
  }
  cnt = min((int)cycle.size(), S - s);
  vector<vector<int> > opt;
  for (int i = cnt; i < cycle.size(); i++) opt.push_back(cycle[i]);
  if (cnt >= 2) {
    opt.push_back(vector<int>());
    for (int i = 0; i < cnt; i++)
      opt.back().insert(opt.back().end(), cycle[i].begin(), cycle[i].end());
    opt.push_back(vector<int>());
    for (int i = 0; i < cnt; i++) opt.back().push_back(cycle[i].front());
    reverse(opt.back().begin(), opt.back().end());
  } else {
    for (int i = 0; i < cnt; i++) opt.push_back(cycle[i]);
  }
  printf("%u\n", opt.size());
  for (auto &x : opt) {
    printf("%u\n", x.size());
    for (auto &y : x) printf("%d ", y);
    puts("");
  }
  return 0;
}