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 M = 0, fst[666666], vb[666666], nxt[666666], vc[666666];
void ad_de(int a, int b, int c) {
  ++M;
  nxt[M] = fst[a];
  fst[a] = M;
  vb[M] = b;
  vc[M] = c;
}
void adde(int a, int b, int c) {
  ad_de(a, b, c);
  ad_de(b, a, c);
}
map<int, int> vis;
int n, s, a[666666], b[666666], id, vv[666666], ed;
vector<int> cur, vs[666666];
int vn;
void dfs(int x) {
  for (int& e = fst[x]; e; e = nxt[e])
    if (!vv[vc[e]]) {
      int c = vc[e];
      vv[vc[e]] = 1;
      dfs(vb[e]);
      cur.push_back(c);
    }
}
int main() {
  scanf("%d%d", &n, &s);
  for (int i = 1; i <= n; ++i) scanf("%d", a + i), vis[a[i]] = 1;
  for (auto& t : vis) t.second = ++id;
  for (int i = 1; i <= n; ++i) b[i] = a[i] = vis[a[i]];
  sort(b + 1, b + 1 + n);
  for (int i = 1; i <= n; ++i)
    if (a[i] != b[i]) ++ed, ad_de(a[i], b[i], i);
  if (ed > s) {
    puts("-1");
    return 0;
  }
  for (int i = 1; i <= id; ++i)
    if (fst[i]) {
      dfs(i);
      vs[++vn] = cur;
      cur.clear();
    }
  int tj = min(vn, s - ed);
  if (tj <= 2) tj = 0;
  vector<int> ta, tb;
  for (int i = 1; i <= tj; ++i) {
    ta.insert(ta.end(), vs[vn].begin(), vs[vn].end());
    tb.push_back(vs[vn--].back());
  }
  if (tj) {
    reverse(tb.begin(), tb.end());
    vs[++vn] = tb;
    vs[++vn] = ta;
  }
  printf("%d\n", vn);
  for (int i = 1; i <= vn; ++i) {
    printf("%d\n", int(vs[i].size()));
    for (auto g : vs[i]) printf("%d ", g);
    puts("");
  }
}