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>
template <class T>
bool chmax(T &a, const T &b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
bool chmin(T &a, const T &b) {
  if (a > b) {
    a = b;
    return 1;
  }
  return 0;
}
using namespace std;
template <class A, class B>
ostream &operator<<(ostream &os, const pair<A, B> &p) {
  return os << "(" << p.first << "," << p.second << ")";
}
template <class A, class B>
istream &operator>>(istream &is, pair<A, B> &p) {
  return is >> p.first >> p.second;
}
int main(void) {
  cin.tie(0);
  ios::sync_with_stdio(false);
  int N;
  cin >> N;
  vector<int> a(N);
  for (int i = 0; i < (int)(N); i++) cin >> a[i];
  vector<vector<int> > ans;
  while (true) {
    bool ok = true;
    for (int i = 0; i < (int)(N); i++)
      if (a[i] != i + 1) ok = false;
    if (ok) break;
    {
      vector<vector<int> > vec;
      vec.push_back(vector<int>());
      vec.back().push_back(a[0]);
      for (int i = 0; i < (int)(N - 1); i++) {
        if (a[i + 1] - a[i] != 1) {
          vec.push_back(vector<int>());
        }
        vec.back().push_back(a[i + 1]);
      }
      int K = vec.size();
      {
        int l = -1, r = -1;
        for (int i = 0; i < (int)(K); i++) {
          for (int j = (int)(i + 1); j <= (int)(K - 1); j++) {
            int u = vec[i].front();
            int v = vec[j].back();
            if (u - v == 1) {
              l = i;
              r = j - 1;
              break;
            }
          }
          if (l != -1) break;
        }
        vector<vector<int> > tmp;
        tmp.push_back(vector<int>());
        for (int i = 0; i < (int)(K); i++) {
          for (auto e : vec[i]) {
            tmp.back().push_back(e);
          }
          if (i < K - 1 && !(l <= i && i < r)) tmp.push_back(vector<int>());
        }
        swap(vec, tmp);
        K = vec.size();
      }
      {
        ans.push_back(vector<int>());
        for (int i = 0; i < (int)(K); i++) ans.back().push_back(vec[i].size());
      }
      vector<int> tmp;
      for (int i = (int)(K - 1); i >= (int)(0); i--) {
        for (auto e : vec[i]) {
          tmp.push_back(e);
        }
      }
      swap(tmp, a);
    }
  }
  cout << ans.size() << "\n";
  for (auto v : ans) {
    int M = v.size();
    cout << M << " ";
    for (int i = 0; i < (int)(M); i++) cout << v[i] << " \n"[i == M - 1];
  }
  return 0;
}