You are given a deck of n cards numbered from 1 to n (not necessarily in this order in the deck). You have to sort the deck by repeating the following operation. 

  * Choose 2 ≤ k ≤ n and split the deck in k nonempty contiguous parts D_1, D_2,..., D_k (D_1 contains the first |D_1| cards of the deck, D_2 contains the following |D_2| cards and so on). Then reverse the order of the parts, transforming the deck into D_k, D_{k-1}, ..., D_2, D_1 (so, the first |D_k| cards of the new deck are D_k, the following |D_{k-1}| cards are D_{k-1} and so on). The internal order of each packet of cards D_i is unchanged by the operation. 



You have to obtain a sorted deck (i.e., a deck where the first card is 1, the second is 2 and so on) performing at most n operations. It can be proven that it is always possible to sort the deck performing at most n operations.

Examples of operation: The following are three examples of valid operations (on three decks with different sizes). 

  * If the deck is [3 6 2 1 4 5 7] (so 3 is the first card and 7 is the last card), we may apply the operation with k=4 and D_1=[3 6], D_2=[2 1 4], D_3=[5], D_4=[7]. Doing so, the deck becomes [7 5 2 1 4 3 6]. 
  * If the deck is [3 1 2], we may apply the operation with k=3 and D_1=[3], D_2=[1], D_3=[2]. Doing so, the deck becomes [2 1 3]. 
  * If the deck is [5 1 2 4 3 6], we may apply the operation with k=2 and D_1=[5 1], D_2=[2 4 3 6]. Doing so, the deck becomes [2 4 3 6 5 1]. 

Input

The first line of the input contains one integer n (1≤ n≤ 52) — the number of cards in the deck.

The second line contains n integers c_1, c_2, ..., c_n — the cards in the deck. The first card is c_1, the second is c_2 and so on.

It is guaranteed that for all i=1,...,n there is exactly one j∈\{1,...,n\} such that c_j = i.

Output

On the first line, print the number q of operations you perform (it must hold 0≤ q≤ n).

Then, print q lines, each describing one operation.

To describe an operation, print on a single line the number k of parts you are going to split the deck in, followed by the size of the k parts: |D_1|, |D_2|, ... , |D_k|. 

It must hold 2≤ k≤ n, and |D_i|≥ 1 for all i=1,...,k, and |D_1|+|D_2|+⋅⋅⋅ + |D_k| = n.

It can be proven that it is always possible to sort the deck performing at most n operations. If there are several ways to sort the deck you can output any of them.

Examples

Input


4
3 1 2 4


Output


2
3 1 2 1
2 1 3


Input


6
6 5 4 3 2 1


Output


1
6 1 1 1 1 1 1


Input


1
1


Output


0

Note

Explanation of the first testcase: Initially the deck is [3 1 2 4]. 

  * The first operation splits the deck as [(3) (1 2) (4)] and then transforms it into [4 1 2 3]. 
  * The second operation splits the deck as [(4) (1 2 3)] and then transforms it into [1 2 3 4]. 

Explanation of the second testcase: Initially the deck is [6 5 4 3 2 1]. 

  * The first (and only) operation splits the deck as [(6) (5) (4) (3) (2) (1)] and then transforms it into [1 2 3 4 5 6]. 
#include <bits/stdc++.h>
using namespace std;
long long int Lpow(long long int a, long long int b) {
  long long int ans = 1;
  while (b > 0) {
    if (b % 2) ans = (ans * a);
    a = (a * a);
    b = b / 2;
  }
  return ans;
}
long long int Lpow(long long int a, long long int b, long long int M) {
  long long int ans = 1;
  while (b > 0) {
    if (b % 2) ans = (ans * a) % M;
    a = (a * a) % M;
    b = b / 2;
  }
  return ans;
}
int arr[55];
vector<vector<int>> ans(100);
int idx = 0;
void chng(vector<int> y, int n) {
  int brr[n];
  reverse(y.begin(), y.end());
  int last = n;
  int curr_a = n - 1;
  int curr_b = 0;
  for (int i = 0; i < y.size(); i++) {
    int str = curr_a - y[i] + 1;
    for (int j = 0; j < y[i]; j++) {
      brr[curr_b] = arr[str];
      str++;
      curr_b++;
    }
    curr_a = curr_a - y[i];
  }
  for (int i = 0; i < n; i++) arr[i] = brr[i];
}
void fun(int done, int n) {
  if (n - 2 * done <= 1) return;
  int min_loc = -1;
  int mi = 2147483647;
  int max_loc = -1;
  int ma = 0;
  for (int i = done; i < n - done; i++) {
    if (arr[i] < mi) {
      mi = arr[i];
      min_loc = i;
    }
    if (arr[i] > ma) {
      ma = arr[i];
      max_loc = i;
    }
  }
  if (max_loc < min_loc) {
    for (int i = 0; i < done; i++) ans[idx].push_back(1);
    ans[idx].push_back(max_loc - done + 1);
    if (min_loc - max_loc - 1 > 0) ans[idx].push_back(min_loc - max_loc - 1);
    ans[idx].push_back(n - min_loc - done);
    for (int i = 0; i < done; i++) ans[idx].push_back(1);
    chng(ans[idx], n);
    if (done > 0) {
      idx++;
      for (int i = 0; i < done; i++) ans[idx].push_back(1);
      if (n - 2 * done > 0) ans[idx].push_back(n - 2 * done);
      for (int i = 0; i < done; i++) ans[idx].push_back(1);
      chng(ans[idx], n);
    }
  } else {
    for (int i = 0; i < done; i++) ans[idx].push_back(1);
    ans[idx].push_back(min_loc - done + 1);
    if (max_loc - min_loc - 1 > 0) ans[idx].push_back(max_loc - min_loc - 1);
    ans[idx].push_back(n - max_loc - done);
    for (int i = 0; i < done; i++) ans[idx].push_back(1);
    chng(ans[idx], n);
    idx++;
    for (int i = 0; i < done + 1; i++) ans[idx].push_back(1);
    if (n - 2 * (done + 1) > 0) ans[idx].push_back(n - 2 * (done + 1));
    for (int i = 0; i < done + 1; i++) ans[idx].push_back(1);
    chng(ans[idx], n);
  }
  done++;
  idx++;
  fun(done, n);
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) cin >> arr[i];
  fun(0, n);
  cout << idx << '\n';
  for (int i = 0; i < idx; i++) {
    cout << ans[i].size() << " ";
    for (auto x : ans[i]) cout << x << " ";
    cout << '\n';
  }
  return 0;
}
