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>
inline int rint() {
  int x = 0;
  char s = getchar();
  for (; s < '0' || '9' < s; s = getchar())
    ;
  for (; '0' <= s && s <= '9'; s = getchar()) x = x * 10 + (s ^ '0');
  return x;
}
template <typename Tp>
inline void wint(Tp x) {
  if (x < 0) putchar('-'), x = -x;
  if (9 < x) wint(x / 10);
  putchar(x % 10 ^ '0');
}
const int MAXN = 2e5;
int n, s, ecnt = 1, a[MAXN + 5], tmp[MAXN + 5], rk[MAXN + 5],
          head[MAXN * 2 + 5];
bool vis[MAXN + 5];
struct Edge {
  int to, nxt;
} graph[MAXN * 2 + 5];
std::vector<std::vector<int> > ans;
inline void link(const int s, const int t) {
  graph[++ecnt].to = t, graph[ecnt].nxt = head[s];
  head[s] = ecnt;
}
inline void findEC(const int u, std::vector<int>& res) {
  if (u <= n) vis[u] = true;
  for (int &i = head[u], v; i;) {
    v = graph[i].to, i = graph[i].nxt;
    findEC(v, res);
  }
  if (u <= n) res.push_back(u);
}
inline void printEC(const std::vector<int>& path) {
  for (int i = path.size() - 1; ~i; --i) {
    wint(path[i]);
    if (i) putchar(' ');
  }
}
int main() {
  n = rint(), s = rint();
  for (int i = 1; i <= n; ++i) a[i] = tmp[i] = rint();
  std::sort(tmp + 1, tmp + n + 1);
  int mx = std::unique(tmp + 1, tmp + n + 1) - tmp - 1;
  for (int i = 1; i <= n; ++i) {
    a[i] = std::lower_bound(tmp + 1, tmp + mx + 1, a[i]) - tmp;
    ++rk[a[i]];
  }
  for (int i = 2; i <= mx; ++i) rk[i] += rk[i - 1];
  int cnt = 0;
  for (int i = 1; i <= n; ++i) {
    if (rk[a[i] - 1] < i && i <= rk[a[i]]) continue;
    link(i, n + a[i]), ++cnt;
  }
  if ((s -= cnt) < 0) return puts("-1"), 0;
  for (int i = 1; i <= mx; ++i) {
    for (int j = rk[i - 1] + 1; j <= rk[i]; ++j) {
      if (a[j] == i) continue;
      link(n + i, j);
    }
  }
  std::vector<int> tmp;
  for (int i = 1; i <= n; ++i) {
    if (!vis[i]) {
      tmp.clear();
      findEC(i, tmp);
      if (tmp.size() == 1) continue;
      tmp.pop_back();
      ans.push_back(tmp);
    }
  }
  int sz = ans.size();
  if (s > sz) s = sz;
  int norid = 0;
  if (s > 2) sz += 2 - s;
  wint(sz), putchar('\n');
  if (s > 2) {
    int firs = 0;
    for (int i = 0; i < s; ++i) firs += ans[i].size();
    wint(firs), putchar('\n');
    for (int i = 0; i < s; ++i) {
      printEC(ans[i]), putchar(i + 1 < s ? ' ' : '\n');
    }
    wint(s), putchar('\n');
    wint(ans[0].back());
    for (int i = s - 1; i; --i) {
      putchar(' '), wint(ans[i].back());
    }
    putchar('\n');
    norid = s;
  }
  for (; norid ^ ans.size(); ++norid) {
    wint(ans[norid].size()), putchar('\n');
    printEC(ans[norid]), putchar('\n');
  }
  return 0;
}