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;
int n, s, t, ans, a[N];
int p[N], fa[N];
pair<int, int> b[N];
int get(int x) { return x == fa[x] ? x : fa[x] = get(fa[x]); }
bool vis[N];
vector<int> c[N];
void dfs(int x) {
  vis[x] = 1;
  c[t].push_back(x);
  if (!vis[p[x]]) dfs(p[x]);
}
int main() {
  scanf("%d%d", &n, &s);
  for (int i = (int)(1); i <= (int)(n); i++) {
    scanf("%d", &a[i]);
    b[i] = pair<int, int>(a[i], i);
  }
  sort(b + 1, b + n + 1);
  for (int i = (int)(1); i <= (int)(n); i++) p[b[i].second] = i;
  for (int i = (int)(1); i <= (int)(n); i++)
    if (a[i] == b[i].first && p[i] != i) {
      p[b[i].second] = p[i];
      b[p[i]].second = b[i].second;
      p[i] = b[i].second = i;
    }
  for (int i = (int)(1); i <= (int)(n); i++) fa[i] = i;
  for (int i = (int)(1); i <= (int)(n); i++) fa[get(i)] = get(p[i]);
  int las = -1;
  for (int i = (int)(0); i <= (int)(n); i++) {
    if (p[b[i].second] == b[i].second) continue;
    if (las >= 0 && a[las] == a[b[i].second]) {
      int x = las, y = b[i].second;
      int X = get(x), Y = get(y);
      if (X != Y) fa[X] = Y, swap(p[x], p[y]);
    }
    las = b[i].second;
  }
  for (int i = (int)(1); i <= (int)(n); i++)
    if (!vis[i] && p[i] != i) ++t, dfs(i);
  for (int i = (int)(1); i <= (int)(t); i++) ans += c[i].size();
  if (ans > s) return puts("-1"), 0;
  s = min(s - ans, t);
  if (s <= 1) {
    printf("%d\n", t);
    for (int i = (int)(1); i <= (int)(t); i++) {
      printf("%d\n", c[i].size());
      for (int j = (int)(0); j <= (int)(c[i].size() - 1); j++)
        printf("%d ", c[i][j]);
      puts("");
    }
    return 0;
  }
  printf("%d\n", t - s + 2);
  for (int i = (int)(1); i <= (int)(t - s); i++) {
    printf("%d\n", c[i + s].size());
    for (int j = (int)(0); j <= (int)(c[i + s].size() - 1); j++)
      printf("%d ", c[i + s][j]);
    puts("");
    ans -= c[i + s].size();
  }
  printf("%d\n", ans);
  for (int i = (int)(1); i <= (int)(s); i++)
    for (int j = (int)(0); j <= (int)(c[i].size() - 1); j++)
      printf("%d ", c[i][j]);
  printf("\n%d\n", s);
  for (int i = (int)(s); i >= (int)(1); i--) printf("%d ", c[i][0]);
}