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 = 2e5 + 5;
int n, S;
int a[N], p[N], q[N];
vector<vector<int>> cir;
int uf[N];
int find(int x) { return (uf[x] == x) ? (x) : (uf[x] = find(uf[x])); }
void adjust() {
  for (int i = 1; i <= n; i++) {
    uf[i] = i;
  }
  pair<int, int>* b = new pair<int, int>[(n + 1)];
  for (int i = 1; i <= n; i++) b[i] = pair<int, int>(a[i], i);
  sort(b + 1, b + n + 1);
  for (int i = 1, j = 1; i <= n; i++, i = j) {
    while (j <= n && b[i].first == b[j].first) j++;
    for (int k = i, x; k < j; k++) {
      x = b[k].second;
      if (x >= i && x < j) {
        q[x] = x;
      }
    }
    for (int k = i, cur = i; k < j; k++) {
      if (!q[k]) {
        while (cur < j && q[b[cur].second] == b[cur].second) cur++;
        q[k] = b[cur++].second;
      }
    }
  }
  for (int i = 1; i <= n; i++) {
    p[q[i]] = i;
  }
  for (int i = 1; i <= n; i++) {
    uf[find(i)] = find(p[i]);
  }
  for (int i = 1, j = 1, ls; i <= n; i = j) {
    while (j <= n && b[i].first == b[j].first) j++;
    for (ls = i; ls < j && q[ls] == ls; ls++)
      ;
    for (int k = ls + 1; k < j; k++) {
      if (q[k] != k && find(q[k]) != find(q[ls])) {
        uf[find(q[k])] = find(ls);
        uf[find(q[ls])] = find(k);
        swap(p[q[ls]], p[q[k]]);
        swap(q[ls], q[k]);
        ls = k;
      }
    }
  }
}
bool vis[N];
int find_circle(int x) {
  if (p[x] == x) {
    return 0;
  }
  cir.push_back(vector<int>());
  do {
    vis[x] = true;
    cir.back().push_back(x);
    x = p[x];
  } while (!vis[x]);
  return cir.back().size();
}
int main() {
  scanf("%d%d", &n, &S);
  for (int i = 1; i <= n; i++) {
    scanf("%d", a + i);
  }
  adjust();
  int t = 0, m = 0;
  for (int i = 1, x; i <= n; i++) {
    if (!vis[i]) {
      x = find_circle(i);
      t += x;
      m += (x != 0);
    }
  }
  if (t > S) {
    puts("-1");
    return 0;
  }
  if (m <= 2 || t + 2 >= S) {
    printf("%d\n", m);
    for (int i = 0; i < m; i++) {
      printf("%d\n", (signed)cir[i].size());
      for (auto& e : cir[i]) {
        printf("%d ", e);
      }
      putchar('\n');
    }
  } else {
    int c = min(m, S - t);
    printf("%d\n", 2 + m - c);
    int sum = 0;
    for (int i = 0; i < c; i++) sum += cir[i].size();
    printf("%d\n", sum);
    for (int i = 0; i < c; i++) {
      for (auto& e : cir[i]) {
        printf("%d ", e);
      }
    }
    putchar('\n');
    printf("%d\n%d", c, cir[0][0]);
    for (int i = c - 1; i; i--) printf(" %d", cir[i][0]);
    putchar('\n');
    for (int i = c; i < m; i++) {
      printf("%d\n", (signed)cir[i].size());
      for (auto& e : cir[i]) {
        printf("%d ", e);
      }
      putchar('\n');
    }
  }
  return 0;
}