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[200020], b[200020], p[200020];
bool cmp(int i, int j) { return a[i] < a[j]; }
int rt[200020];
int findrt(int x) {
  if (rt[x] != x) rt[x] = findrt(rt[x]);
  return rt[x];
}
int get(int x) { return lower_bound(b + 1, b + n + 1, x) - b; }
map<int, int> S;
int c[200020];
bool vis[200020];
int tot;
vector<int> ans[200020];
int q[200020];
int main() {
  cin >> n >> s;
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]), b[i] = a[i];
  sort(b + 1, b + n + 1);
  int cnt = n;
  for (int i = 1; i <= n; i++)
    if (a[i] == b[i]) q[i] = i, cnt--;
  if (cnt > s) {
    puts("-1");
    return 0;
  }
  vector<int> v;
  for (int i = 1; i <= n; i++)
    if (!q[i]) v.push_back(i);
  sort(v.begin(), v.end(), cmp);
  for (int i = 1, j = 0; i <= n; i++) {
    if (q[i]) continue;
    q[i] = v[j];
    j++;
  }
  for (int i = 1; i <= n; i++) p[q[i]] = i;
  for (int i = 1; i <= n; i++) assert(a[i] == b[p[i]]);
  for (int i = 1; i <= n; i++) rt[i] = i;
  for (int i = 1; i <= n; i++) {
    if (p[i] == i) continue;
    int l = findrt(i), r = findrt(p[i]);
    if (l == r) continue;
    rt[l] = r;
  }
  for (int i = 1; i <= n; i++) {
    if (p[i] == i) continue;
    if (!S.count(a[i])) {
      S[a[i]] = i;
      continue;
    }
    int l = S[a[i]];
    int fl = findrt(l), fr = findrt(i);
    if (fl == fr) continue;
    rt[fr] = fl;
    swap(p[l], p[i]);
  }
  for (int i = 1; i <= n; i++) {
    if (p[i] == i) {
      vis[i] = 1;
      continue;
    }
    if (vis[i]) continue;
    tot++;
    vector<int> &cur = ans[tot];
    int now = i;
    while (1) {
      vis[now] = 1;
      cur.push_back(now);
      now = p[now];
      if (now == i) break;
    }
  }
  int k = s - cnt;
  k = min(tot, k);
  if (k >= 3) {
    vector<int> cur;
    for (int i = 0; i < k; i++) cur.push_back(ans[tot - i][0]);
    vector<int> &to = ans[tot - k + 1];
    for (int i = k - 2; i >= 0; i--) {
      for (int x : ans[tot - i]) to.push_back(x);
    }
    ans[tot - k + 2] = cur;
    tot -= k - 2;
  }
  cout << tot << endl;
  for (int i = 1; i <= tot; i++) {
    cout << ans[i].size() << endl;
    for (int x : ans[i]) printf("%d ", x);
    puts("");
  }
}