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 std::lower_bound;
using std::max;
using std::min;
using std::random_shuffle;
using std::reverse;
using std::sort;
using std::swap;
using std::unique;
using std::upper_bound;
using std::vector;
void open(const char *s) {}
int rd() {
  int s = 0, c, b = 0;
  while (((c = getchar()) < '0' || c > '9') && c != '-')
    ;
  if (c == '-') {
    c = getchar();
    b = 1;
  }
  do {
    s = s * 10 + c - '0';
  } while ((c = getchar()) >= '0' && c <= '9');
  return b ? -s : s;
}
void put(int x) {
  if (!x) {
    putchar('0');
    return;
  }
  static int c[20];
  int t = 0;
  while (x) {
    c[++t] = x % 10;
    x /= 10;
  }
  while (t) putchar(c[t--] + '0');
}
int upmin(int &a, int b) {
  if (b < a) {
    a = b;
    return 1;
  }
  return 0;
}
int upmax(int &a, int b) {
  if (b > a) {
    a = b;
    return 1;
  }
  return 0;
}
const int N = 200010;
int n, m;
int a[N], b[N], c[N];
int d[N];
vector<std::pair<int, int> > g[N];
int cnt;
vector<int> s[N];
void dfs(int x) {
  while (!g[x].empty()) {
    auto v = g[x].back();
    g[x].pop_back();
    dfs(v.first);
    s[cnt].push_back(v.second);
  }
}
int main() {
  open("cf1012e");
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    b[i] = a[i];
    c[i] = a[i];
  }
  sort(b + 1, b + n + 1);
  sort(c + 1, c + n + 1);
  int t = unique(c + 1, c + n + 1) - c - 1;
  for (int i = 1; i <= n; i++) {
    a[i] = lower_bound(c + 1, c + t + 1, a[i]) - c;
    b[i] = lower_bound(c + 1, c + t + 1, b[i]) - c;
  }
  for (int i = 1; i <= n; i++)
    if (a[i] != b[i]) g[a[i]].push_back(std::pair<int, int>(b[i], i));
  for (int i = 1; i <= t; i++) {
    cnt++;
    dfs(i);
    if (s[cnt].empty()) cnt--;
  }
  int sum = 0;
  for (int i = 1; i <= cnt; i++) sum += s[i].size();
  if (sum > m) {
    printf("-1\n");
    return 0;
  }
  if (cnt == 0) {
    printf("0\n");
    return 0;
  }
  if (cnt == 1) {
    printf("1\n%d\n", sum);
    for (auto v : s[1]) printf("%d ", v);
    printf("\n");
    return 0;
  }
  for (int i = cnt; i >= 2; i--)
    if (i + sum <= m) {
      printf("%d\n", cnt - i + 2);
      int sum1 = 0;
      for (int j = 1; j <= i; j++) sum1 += s[j].size();
      printf("%d\n", sum1);
      for (int j = 1; j <= i; j++)
        for (auto v : s[j]) printf("%d ", v);
      printf("\n");
      printf("%d\n", i);
      for (int j = i; j >= 1; j--) printf("%d ", s[j].front());
      for (int j = i + 1; j <= cnt; j++) {
        printf("%d\n", (int)s[j].size());
        for (auto v : s[j]) printf("%d ", v);
        printf("\n");
      }
      return 0;
    }
  printf("%d\n", cnt);
  for (int i = 1; i <= cnt; i++) {
    printf("%d\n", (int)s[i].size());
    for (auto v : s[i]) printf("%d ", v);
    printf("\n");
  }
  return 0;
}