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 = 520233;
int n, s, a[N], b[N];
int fa[N];
int Find(int x) { return (fa[x] == x) ? x : (fa[x] = Find(fa[x])); }
map<int, int> mp;
bool Merge(int x, int y) {
  if (Find(x) != Find(y)) {
    fa[Find(x)] = Find(y);
    return true;
  }
  return false;
}
int p[N], cnt;
vector<int> v[N];
bool vis[N];
void Dfs(int u) {
  v[cnt].emplace_back(u);
  vis[u] = true;
  if (!vis[p[u]]) Dfs(p[u]);
}
map<int, vector<int> > lst;
map<int, int> rec;
int main() {
  scanf("%d%d", &n, &s);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", a + i);
    b[i] = a[i];
    fa[i] = i;
  }
  sort(b + 1, b + n + 1);
  int least = 0;
  for (int i = 1; i <= n; ++i)
    if (a[i] != b[i]) {
      ++least;
      lst[b[i]].emplace_back(i);
    }
  if (least > s) {
    puts("-1");
    return 0;
  } else if (!least) {
    puts("0");
    return 0;
  }
  for (int i = 1; i <= n; ++i)
    if (a[i] != b[i]) {
      vector<int>& v = lst[a[i]];
      p[i] = v.back();
      Merge(i, p[i]);
      v.pop_back();
    }
  for (int i = 1; i <= n; ++i)
    if (a[i] != b[i]) {
      int& t = rec[a[i]];
      if (t && Merge(i, t)) swap(p[i], p[t]);
      t = i;
    }
  for (int i = 1; i <= n; ++i)
    if (a[i] != b[i] && !vis[i]) {
      ++cnt;
      Dfs(i);
    }
  int q = s - least;
  if (q > 1)
    --q;
  else
    q = 0;
  q = min(q, cnt - 1);
  if (q) {
    printf("%d\n", cnt - q + 1);
    int sm = 0;
    for (int i = 1; i <= q + 1; ++i) sm += v[i].size();
    printf("%d\n", sm);
    for (int i = 1; i <= q + 1; ++i)
      for (int j : v[i]) printf("%d ", j);
    putchar('\n');
    printf("%d\n", q + 1);
    for (int i = q + 1; i; --i) printf("%d ", v[i][0]);
    putchar('\n');
    for (int i = q + 2; i <= cnt; ++i) {
      printf("%d\n", (int)v[i].size());
      for (int j : v[i]) printf("%d ", j);
      putchar('\n');
    }
  } else {
    printf("%d\n", cnt);
    for (int i = 1; i <= cnt; ++i) {
      printf("%d\n", (int)v[i].size());
      for (int j : v[i]) printf("%d ", j);
      putchar('\n');
    }
  }
  return 0;
}