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 = 200005;
map<int, int> MP;
struct node {
  int a, b;
  inline node() {}
  inline node(int a, int b) : a(a), b(b) {}
  bool operator<(node x) const { return a < x.a; }
} nod[N];
int a[N], nxt[N], fa[N], rt[N], sz[N], b[N];
int find(int v) { return fa[v] == 0 ? v : fa[v] = find(fa[v]); }
int main() {
  int n, m, i, t = 0, x, y, sum;
  scanf("%d%d", &n, &m);
  for (i = 1; i <= n; ++i) scanf("%d", &a[i]);
  for (i = 1; i <= n; ++i) b[i] = a[i];
  sort(b + 1, b + n + 1);
  for (i = 1; i <= n; ++i)
    if (a[i] != b[i]) {
      a[++t] = a[i];
      b[t] = i;
      nod[t] = node(a[t], t);
    }
  if (t > m) {
    printf("-1");
    return 0;
  }
  if (t == 0) {
    printf("0");
    return 0;
  }
  sort(nod + 1, nod + (n = t) + 1);
  for (i = 1; i <= n; ++i) nxt[nod[i].b] = i;
  for (i = 1; i <= n; ++i)
    if (fa[i] == 0)
      for (x = nxt[i]; x != i; x = nxt[x]) fa[x] = i;
  for (i = 1; i <= n; ++i) {
    t = MP[a[i]];
    if (t != 0) {
      x = find(i);
      y = find(t);
      if (x != y) {
        fa[x] = y;
        swap(nxt[i], nxt[t]);
      }
    }
    MP[a[i]] = i;
  }
  t = 0;
  for (i = 1; i <= n; ++i)
    if (fa[i] == 0) rt[++t] = i;
  for (i = 1; i <= n; ++i) ++sz[find(i)];
  if (t == 1 || m <= n + 2) {
    printf("%d", t);
    for (i = 1; i <= t; ++i) {
      x = rt[i];
      printf("\n%d\n%d", sz[x], b[x]);
      for (y = nxt[x]; y != x; y = nxt[y]) printf(" %d", b[y]);
    }
  } else {
    m = min(m - n, t);
    sum = 0;
    for (i = 1; i <= m; ++i) sum += sz[rt[i]];
    printf("%d\n%d\n", t - m + 2, sum);
    for (i = 1; i <= m; ++i) {
      x = rt[i];
      printf("%d ", b[x]);
      for (y = nxt[x]; y != x; y = nxt[y]) printf("%d ", b[y]);
    }
    printf("\n%d\n", m);
    for (i = m; i >= 1; --i) printf("%d ", b[rt[i]]);
    for (i = m + 1; i <= t; ++i) {
      x = rt[i];
      printf("\n%d\n%d", sz[x], b[x]);
      for (y = nxt[x]; y != x; y = nxt[y]) printf(" %d", b[y]);
    }
  }
  return 0;
}