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 = 4e5 + 10;
int Begin[N], Next[N], to[N], e;
void add(int u, int v) { to[++e] = v, Next[e] = Begin[u], Begin[u] = e; }
int n, m, k, c;
int A[N], rk[N], st[N], vis[N];
void DFS(int o) {
  vis[o] = true;
  for (int& i = Begin[o]; i;) {
    int u = to[i];
    i = Next[i];
    DFS(u);
    st[++c] = o;
  }
}
bool cmp(int x, int y) { return A[x] < A[y]; }
int L[N], R[N], pos[N];
vector<int> seq[N], cyc[N];
struct Edge {
  int u, v, id;
  bool operator<(const Edge& E) const { return u != E.u ? u < E.u : v < E.v; }
};
multiset<Edge> S;
bool used[N];
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) scanf("%d", &A[i]), rk[i] = i;
  sort(rk + 1, rk + n + 1, cmp);
  for (int i = 1; i <= n; i++) L[i] = A[rk[i]] == A[rk[i - 1]] ? L[i - 1] : i;
  for (int i = n; i >= 1; i--) R[i] = A[rk[i]] == A[rk[i + 1]] ? R[i + 1] : i;
  for (int i = 1; i <= n; i++)
    if (rk[i] >= L[i] && rk[i] <= R[i]) used[rk[i]] = true;
  for (int i = 1; i <= n; i++)
    if (L[i] == i) pos[i] = i;
  int ans = 0;
  for (int i = 1; i <= n; i++)
    if (rk[i] < L[i] || rk[i] > R[i])
      ++ans, add(L[rk[i]], L[i]), S.insert((Edge){L[rk[i]], L[i], rk[i]});
  if (ans > k) {
    puts("-1");
    return 0;
  }
  for (int i = 1; i <= n; i++)
    if (!vis[i] && Begin[i]) {
      c = 0;
      DFS(i);
      ++m;
      for (int j = c; j >= 1; j--) {
        int v = j == 1 ? st[c] : st[j - 1];
        auto it = S.lower_bound((Edge){st[j], v, 0});
        seq[m].push_back(it->id);
        S.erase(it);
      }
    }
  int p = 1;
  c = 0;
  if (k - ans > 2 && m > 2) {
    int pr = min(m, k - ans);
    c = 1;
    for (int i = 1; i <= pr; i++)
      for (int v : seq[i]) cyc[c].push_back(v);
    c = 2;
    for (int i = pr; i >= 1; i--) cyc[c].push_back(seq[i].front());
    p = pr + 1;
  }
  for (int i = p; i <= m; i++) cyc[++c] = seq[i];
  printf("%d\n", c);
  for (int i = 1; i <= c; i++) {
    printf("%lu\n", cyc[i].size());
    for (int v : cyc[i]) printf("%d%c", v, v == cyc[i].back() ? '\n' : ' ');
  }
  return 0;
}