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 = 2e5 + 3;
unordered_map<int, int> mp;
int a[N], b[N], he[N], to[N], ne[N], c;
basic_string<int> w[N];
bool v[N];
void dfs(int x) {
  v[x] = 1;
  for (int i; i = he[x];) he[x] = ne[i], dfs(to[i]), w[c] += i;
}
int main() {
  int t = 0, i, j, k, n, s;
  scanf("%d%d", &n, &s);
  for (i = 1; i <= n; ++i)
    if (scanf("%d", a + i), b[i] = a[i], !mp[a[i]]) mp[a[i]] = ++t;
  for (i = 1, sort(b + 1, b + n + 1); i <= n; ++i)
    if (a[i] != b[i])
      --s, j = mp[a[i]], k = mp[b[i]], ne[i] = he[j], to[i] = k, he[j] = i;
  if (s < 0) puts("-1"), exit(0);
  for (i = 1; i <= t; ++i)
    if (!v[i])
      if (++c, dfs(i), !w[c].size()) --c;
  if (!c) puts("0"), exit(0);
  if ((s = min(s, c)) > 1) {
    printf("%d\n%d\n", c - s + 2, s);
    for (i = s, j = 0; i; --i) printf("%d ", w[i].back()), j += w[i].size();
    for (printf("\n%d\n", j), i = 1; i <= s; ++i)
      for (int o : w[i]) printf("%d ", o);
    puts("");
  } else
    printf("%d\n", c), i = 1;
  for (; i <= c; ++i, puts("")) {
    cout << w[i].size() << '\n';
    for (int o : w[i]) printf("%d ", o);
  }
  return 0;
}