#include <bits/stdc++.h>
using namespace std;
int fail() {
  printf("-1\n");
  return 0;
}
const int MN = 2e5 + 100;
const int MS = MN;
int N, S, a[MN], v[MN], g[MN], V, f;
using i2 = array<int, 2>;
i2 b[MN];
using vi2 = vector<i2>;
vi2 w[MN];
bool u[MN];
using vi = vector<int>;
vi c[MN];
int C;
void dfs(int n, int p = -1) {
  u[n] = true;
  for (; not w[n].empty();) {
    i2 x = w[n].back();
    w[n].pop_back();
    dfs(x[1], x[0]);
  }
  if (p != -1) c[C].push_back(p);
}
void solve(int n) {
  c[C].clear();
  dfs(n);
  assert(not c[C].empty());
  C++;
}
void cmb(vi* s, vi* e) {
  int z = 0;
  for (vi* t = s; t != e; t++) z += t->size();
  printf("%d\n", z);
  for (vi* t = s; t != e; t++)
    for (int i = t->size() - 1; i >= 0; i--) printf("%d ", t->at(i) + 1);
  printf("\n%ld\n", e - s);
  for (vi* t = e;;) {
    if (t == s) break;
    t--;
    printf("%d ", t->back() + 1);
  }
  printf("\n");
}
int main() {
  scanf("%d%d", &N, &S);
  for (int i = 0; i < N; i++) scanf("%d", a + i), v[i] = a[i];
  sort(v, v + N);
  V = -1;
  for (int i = 0; i < N; i++) {
    if (not i or v[i] != v[i - 1]) b[++V][0] = i;
    v[V] = v[i], g[i] = V;
    b[V][1] = i + 1;
  }
  V++;
  f = 0;
  for (int i = 0; i < N; i++)
    f += i < b[a[i] = lower_bound(v, v + V, a[i]) - v][0] or b[a[i]][1] <= i;
  if (f > S) return fail();
  for (int i = 0; i < N; i++)
    if (a[i] != g[i]) w[g[i]].push_back({i, a[i]});
  for (int i = 0; i < V; i++) u[i] = false;
  C = 0;
  for (int i = 0; i < V; i++)
    if (not u[i] and not w[i].empty()) solve(i);
  int x = min(S - f, C);
  if (x < 2) x = 0;
  printf("%d\n", C + (x ? 2 - x : 0));
  if (x) cmb(c, c + x);
  for (int i = x; i < C; i++) {
    printf("%lu\n", c[i].size());
    for (int j = c[i].size() - 1; j >= 0; j--) printf("%d ", c[i][j] + 1);
    printf("\n");
  }
  return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int fail() {
  printf("-1\n");
  return 0;
}
const int MN = 2e5 + 100;
const int MS = MN;
int N, S, a[MN], v[MN], g[MN], V, f;
using i2 = array<int, 2>;
i2 b[MN];
using vi2 = vector<i2>;
vi2 w[MN];
bool u[MN];
using vi = vector<int>;
vi c[MN];
int C;
void dfs(int n, int p = -1) {
  u[n] = true;
  for (; not w[n].empty();) {
    i2 x = w[n].back();
    w[n].pop_back();
    dfs(x[1], x[0]);
  }
  if (p != -1) c[C].push_back(p);
}
void solve(int n) {
  c[C].clear();
  dfs(n);
  assert(not c[C].empty());
  C++;
}
void cmb(vi* s, vi* e) {
  int z = 0;
  for (vi* t = s; t != e; t++) z += t->size();
  printf("%d\n", z);
  for (vi* t = s; t != e; t++)
    for (int i = t->size() - 1; i >= 0; i--) printf("%d ", t->at(i) + 1);
  printf("\n%ld\n", e - s);
  for (vi* t = e;;) {
    if (t == s) break;
    t--;
    printf("%d ", t->back() + 1);
  }
  printf("\n");
}
int main() {
  scanf("%d%d", &N, &S);
  for (int i = 0; i < N; i++) scanf("%d", a + i), v[i] = a[i];
  sort(v, v + N);
  V = -1;
  for (int i = 0; i < N; i++) {
    if (not i or v[i] != v[i - 1]) b[++V][0] = i;
    v[V] = v[i], g[i] = V;
    b[V][1] = i + 1;
  }
  V++;
  f = 0;
  for (int i = 0; i < N; i++)
    f += i < b[a[i] = lower_bound(v, v + V, a[i]) - v][0] or b[a[i]][1] <= i;
  if (f > S) return fail();
  for (int i = 0; i < N; i++)
    if (a[i] != g[i]) w[g[i]].push_back({i, a[i]});
  for (int i = 0; i < V; i++) u[i] = false;
  C = 0;
  for (int i = 0; i < V; i++)
    if (not u[i] and not w[i].empty()) solve(i);
  int x = min(S - f, C);
  if (x < 2) x = 0;
  printf("%d\n", C + (x ? 2 - x : 0));
  if (x) cmb(c, c + x);
  for (int i = x; i < C; i++) {
    printf("%lu\n", c[i].size());
    for (int j = c[i].size() - 1; j >= 0; j--) printf("%d ", c[i][j] + 1);
    printf("\n");
  }
  return 0;
}

```