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 N = 200000;
int n, m;
int a[N + 1];
vector<int> nums;
void discrete() {
  sort(nums.begin(), nums.end());
  nums.resize(unique(nums.begin(), nums.end()) - nums.begin());
  for (int i = 1; i <= n; i++)
    a[i] = lower_bound(nums.begin(), nums.end(), a[i]) - nums.begin();
}
int val_to[N], pos_to[N + 1];
vector<pair<int, int> > nei[N + 1];
vector<bool> ban[N + 1];
vector<int> pa;
vector<vector<int> > ans;
bool vis[N + 1];
int now[N + 1];
void dfs(int x) {
  vis[x] = true;
  for (int &i = now[x]; i < nei[x].size(); i++)
    if (!ban[x][i]) {
      int y = nei[x][i].first, y0 = nei[x][i].second;
      ban[x][i] = true;
      dfs(y);
      pa.push_back(y0);
    }
}
int main() {
  cin >> n >> m;
  for (int i = 1; i <= n; i++) scanf("%d", a + i), nums.push_back(a[i]);
  discrete();
  nums.clear();
  for (int i = 1; i <= n; i++) nums.push_back(a[i]);
  sort(nums.begin(), nums.end());
  for (int i = 0; i < n; i++) {
    if (!val_to[nums[i]]) val_to[nums[i]] = i + 1;
    pos_to[i + 1] = val_to[nums[i]];
  }
  int cnt = 0;
  for (int i = 1; i <= n; i++)
    if (pos_to[i] != val_to[a[i]])
      cnt++, nei[pos_to[i]].push_back(make_pair(val_to[a[i]], i)),
          ban[pos_to[i]].push_back(false);
  if (cnt > m) return puts("-1"), 0;
  for (int i = 1; i <= n; i++)
    if (!vis[i] && nei[i].size()) pa.clear(), dfs(i), ans.push_back(pa);
  if (min(int(ans.size()), m - cnt) > 1) {
    pa.clear();
    for (int i = 0; i + 1 < min(int(ans.size()), m - cnt); i++)
      swap(a[ans[i][0]], a[ans[i + 1][0]]);
    for (int i = 0; i < min(int(ans.size()), m - cnt); i++)
      pa.push_back(ans[i][0]);
    ans.clear();
    ans.push_back(pa);
  } else
    ans.clear();
  for (int i = 1; i <= n; i++)
    nei[i].clear(), ban[i].clear(), vis[i] = now[i] = 0;
  for (int i = 1; i <= n; i++)
    if (pos_to[i] != val_to[a[i]])
      nei[pos_to[i]].push_back(make_pair(val_to[a[i]], i)),
          ban[pos_to[i]].push_back(false);
  for (int i = 1; i <= n; i++)
    if (!vis[i] && nei[i].size()) pa.clear(), dfs(i), ans.push_back(pa);
  cout << ans.size() << "\n";
  for (int i = 0; i < ans.size(); i++) {
    printf("%d\n", int(ans[i].size()));
    for (int j = ans[i].size() - 1; ~j; j--) printf("%d ", ans[i][j]);
    puts("");
  }
  return 0;
}