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;
char buf[25];
const int maxn = 200010;
struct node {
  int w, id;
} c[maxn], d[maxn];
vector<int> v[maxn];
int a[maxn], b[maxn], f[maxn], q[maxn];
bool p[maxn];
int n, m, s, ans;
int read() {
  int x = 0, f = 0;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = 1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = (x << 1) + (x << 3) + (ch ^ 48);
    ch = getchar();
  }
  return f ? -x : x;
}
void write(int x) {
  if (!x) {
    putchar('0');
    return;
  }
  if (x < 0) {
    putchar('-');
    x = -x;
  }
  int cnt = 0;
  while (x) {
    buf[++cnt] = '0' + x % 10;
    x /= 10;
  }
  for (int i = cnt; i >= 1; --i) putchar(buf[i]);
}
bool cmp(node x, node y) { return x.w < y.w; }
int find(int x) {
  if (f[x] != x) f[x] = find(f[x]);
  return f[x];
}
void unions(int x, int y) {
  int l = find(x), r = find(y);
  if (l != r) f[l] = r;
}
int main() {
  n = read();
  s = read();
  for (int i = 1; i <= n; ++i) {
    a[i] = b[i] = read();
  }
  sort(b + 1, b + 1 + n);
  for (int i = 1; i <= n; ++i)
    if (a[i] != b[i]) {
      ++m;
      c[m] = d[m] = (node){a[i], i};
    }
  if (m > s) {
    puts("-1");
    return 0;
  }
  sort(d + 1, d + 1 + m, cmp);
  for (int i = 1; i <= m; ++i) f[c[i].id] = c[i].id;
  for (int i = 1; i <= m; ++i) unions(d[i].id, c[i].id);
  for (int i = 2; i <= m; ++i)
    if (d[i].w == d[i - 1].w && find(d[i].id) != find(d[i - 1].id)) {
      swap(d[i], d[i - 1]);
      unions(d[i].id, d[i - 1].id);
    }
  for (int i = 1; i <= m; ++i) q[d[i].id] = c[i].id, p[d[i].id] = true;
  for (int i = 1; i <= m; ++i) {
    int x = c[i].id;
    if (p[x]) {
      ++ans;
      while (p[x]) {
        p[x] = false;
        v[ans].push_back(x);
        x = q[x];
      }
    }
  }
  int num = min(ans, s - m);
  if (num > 2) {
    write(ans - (num - 2));
    putchar('\n');
    int sum = 0;
    for (int i = 1; i <= num; ++i) sum += v[i].size();
    write(sum);
    putchar('\n');
    for (int i = 1; i <= num; ++i) {
      for (int j = 0; j <= v[i].size() - 1; ++j) {
        write(v[i][j]);
        if (i != num || j != v[i].size() - 1)
          putchar(' ');
        else
          putchar('\n');
      }
    }
    write(num);
    putchar('\n');
    for (int i = num; i >= 1; --i) {
      write(v[i][0]);
      if (i != 1)
        putchar(' ');
      else
        putchar('\n');
    }
    for (int i = num + 1; i <= ans; ++i) {
      write(v[i].size());
      putchar('\n');
      for (int j = 0; j <= v[i].size() - 1; ++j) {
        write(v[i][j]);
        if (j != v[i].size() - 1)
          putchar(' ');
        else
          putchar('\n');
      }
    }
  } else {
    write(ans);
    putchar('\n');
    for (int i = 1; i <= ans; ++i) {
      write(v[i].size());
      putchar('\n');
      for (int j = 0; j <= v[i].size() - 1; ++j) {
        write(v[i][j]);
        if (j != v[i].size() - 1)
          putchar(' ');
        else
          putchar('\n');
      }
    }
  }
  return 0;
}