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>
bool chkmax(T &x, T y) {
  return x < y ? x = y, true : false;
}
template <typename T>
bool chkmin(T &x, T y) {
  return x > y ? x = y, true : false;
}
int readint() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
int n, s;
int a[200005], f[200005], to[200005];
pair<int, int> b[200005];
bool vis[200005];
vector<vector<int> > ans;
int getf(int x) { return x == f[x] ? x : f[x] = getf(f[x]); }
int main() {
  n = readint();
  s = readint();
  for (int i = 1; i <= n; i++) a[i] = readint(), b[i] = make_pair(a[i], i);
  sort(b + 1, b + n + 1);
  for (int i = 1; i <= n; i++) to[b[i].second] = i;
  for (int i = 1; i <= n; i++)
    while (i != to[i] && a[to[i]] == b[to[i]].first)
      swap(b[to[i]].second, b[to[to[i]]].second), swap(to[to[i]], to[i]);
  for (int i = 1; i <= n; i++) f[i] = i;
  int all = n, cnt = 0;
  for (int i = 1; i <= n; i++) {
    if (i == to[i]) all--;
    if (vis[i] || i == to[i]) continue;
    vis[i] = 1;
    cnt++;
    for (int j = to[i]; j != i; j = to[j]) vis[j] = 1, f[j] = i;
  }
  int lst = 0;
  for (int i = 1; i <= n; i++) {
    if (b[i].second == i) continue;
    if (!lst || b[i].first != b[lst].first) {
      lst = i;
      continue;
    }
    if (getf(b[lst].second) != getf(b[i].second)) {
      f[f[b[i].second]] = f[b[lst].second];
      swap(to[b[i].second], to[b[lst].second]);
      swap(b[i].second, b[lst].second);
      cnt--;
    }
  }
  if (all > s) return printf("-1\n"), 0;
  int tmp = s - all, num = 0;
  vector<int> now(0), rem(0);
  for (int i = 1; i <= n; i++) {
    if (i == to[i]) continue;
    if (i != f[i]) continue;
    now.push_back(i);
    for (int j = to[i]; j != i; j = to[j]) now.push_back(j);
    rem.push_back(i);
    num++;
    if (num >= tmp) {
      ans.push_back(now);
      reverse(rem.begin(), rem.end());
      if (rem.size() > 1) ans.push_back(rem);
      now.clear(), rem.clear();
    }
  }
  if (now.size()) ans.push_back(now);
  reverse(rem.begin(), rem.end());
  if (rem.size() > 1) ans.push_back(rem);
  printf("%d\n", ans.size());
  for (auto v : ans) {
    printf("%d\n", v.size());
    for (auto x : v) printf("%d ", x);
    printf("\n");
  }
  return 0;
}