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;
inline int read() {
  int x = 0, neg = 1;
  char c = getchar();
  while (!isdigit(c)) {
    if (c == '-') neg = -1;
    c = getchar();
  }
  while (isdigit(c)) x = x * 10 + c - '0', c = getchar();
  return x * neg;
}
inline int qpow(int x, int e, int _MOD) {
  int ans = 1;
  while (e) {
    if (e & 1) ans = ans * x % _MOD;
    x = x * x % _MOD;
    e >>= 1;
  }
  return ans;
}
int n = read(), s = read(), a[200005], b[200005], c[200005], d[200005], num = 0;
vector<int> g[200005], cyc[200005];
int k = 0;
inline void dfs(int x) {
  while (!g[x].empty()) {
    int y = g[x].back();
    g[x].pop_back();
    dfs(a[y]);
    cyc[k].push_back(y);
  }
}
signed main() {
  for (int i = 1; i <= n; i++) a[i] = read(), c[i] = a[i];
  sort(c + 1, c + n + 1);
  for (int i = 1; i <= n; i++)
    if (c[i] != c[i - 1]) d[++num] = c[i];
  for (int i = 1; i <= n; i++)
    a[i] = lower_bound(d + 1, d + num + 1, a[i]) - d, b[i] = a[i];
  sort(b + 1, b + n + 1);
  int cnt = 0;
  for (int i = 1; i <= n; i++) {
    if (a[i] != b[i]) {
      cnt++;
      g[b[i]].push_back(i);
    }
  }
  if (!cnt) return puts("0"), 0;
  if (cnt > s) return puts("-1"), 0;
  for (int i = 1; i <= num; i++) {
    if (!g[i].empty()) {
      k++;
      dfs(i);
    }
  }
  for (int i = 1; i <= k; i++) reverse(cyc[i].begin(), cyc[i].end());
  if (k == 1) {
    cout << 1 << endl << cyc[1].size() << endl;
    for (__typeof(cyc[1].begin()) it = cyc[1].begin(); it != cyc[1].end(); it++)
      cout << *it << " ";
    return 0;
  }
  if (cnt <= s - k) {
    cout << 2 << endl;
    cout << cnt << endl;
    for (int i = 1; i <= k; i++) {
      for (__typeof(cyc[i].begin()) it = cyc[i].begin(); it != cyc[i].end();
           it++)
        cout << *it << " ";
    }
    puts("");
    cout << k << endl;
    for (int i = k; i >= 1; i--) cout << cyc[i][0] << " ";
    puts("");
    return 0;
  } else {
    int t = cnt - (s - k);
    if (t == k)
      cout << t << endl;
    else if (t == k - 1)
      cout << t + 1 << endl;
    else
      cout << t + 2 << endl;
    for (int i = 1; i <= t; i++) {
      cout << cyc[i].size() << endl;
      for (__typeof(cyc[i].begin()) it = cyc[i].begin(); it != cyc[i].end();
           it++)
        cout << *it << " ";
      puts("");
    }
    int sum = 0;
    if (t != k) {
      for (int i = t + 1; i <= k; i++) sum += cyc[i].size();
      cout << sum << endl;
      for (int i = t + 1; i <= k; i++)
        for (__typeof(cyc[i].begin()) it = cyc[i].begin(); it != cyc[i].end();
             it++)
          cout << *it << " ";
      if (t != k - 1) {
        puts("");
        cout << k - t << endl;
        for (int i = k; i >= t + 1; i--) cout << cyc[i][0] << " ";
      }
    }
  }
  return 0;
}