Problem: 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]. 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int N;
int C[55];
vector<vector<int>> R;
void hoge(vector<int> A) {
  vector<int> RR;
  vector<vector<int>> B;
  int nex = 0;
  for (auto& a : A)
    if (a) {
      RR.push_back(a);
      B.push_back({});
      while (a--) B.back().push_back(C[nex++]);
    }
  assert(nex == N);
  R.push_back(RR);
  reverse((B.begin()), (B.end()));
  nex = 0;
  for (auto& b : B) {
    for (auto& c : b) C[nex++] = c;
  }
}
void solve() {
  int i, j, k, l, r, x, y;
  string s;
  cin >> N;
  for (i = 0; i < (N); i++) cin >> C[i], C[i]--;
  while (1) {
    int P[55] = {};
    vector<vector<int>> V;
    for (i = 0; i < (N); i++) {
      if (V.empty() || C[i - 1] + 1 != C[i]) V.push_back({});
      V.back().push_back(C[i]);
    }
    if (V.size() == 1) break;
    for (y = 0; y < (V.size()); y++)
      for (x = 0; x < (y); x++) {
        if (V[x][0] == V[y].back() + 1) goto out;
      }
  out:
    int D[4] = {};
    for (i = 0; i < (V.size()); i++) {
      if (i < x)
        D[0] += V[i].size();
      else if (i < y)
        D[1] += V[i].size();
      else if (i == y)
        D[2] += V[i].size();
      else
        D[3] += V[i].size();
    }
    hoge({D[0], D[1], D[2], D[3]});
  }
  for (i = 0; i < (N); i++) assert(C[i] == i);
  cout << R.size() << endl;
  for (auto& r : R) {
    cout << r.size();
    for (auto& v : r) cout << " " << v;
    cout << endl;
  }
}
int main(int argc, char** argv) {
  string s;
  int i;
  if (argc == 1) ios::sync_with_stdio(false), cin.tie(0);
  for (i = 0; i < (argc - 1); i++) s += argv[i + 1], s += '\n';
  for (i = 0; i < (s.size()); i++) ungetc(s[s.size() - 1 - i], stdin);
  cout.tie(0);
  solve();
  return 0;
}