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;
char nl = '\n';
using Int = long long;
long long re[53];
void solve() {
  vector<string> answers;
  long long n;
  cin >> n;
  for (long long i = 1; i <= n; i++) {
    cin >> re[i];
  }
  bool reversed = false;
  for (long long i = 1; i <= n; i++) {
    if (re[i] == i && !reversed || re[n - i + 1] == i && reversed) continue;
    vector<long long> temp;
    vector<vector<long long>> r;
    for (long long j = 1; j <= n; j++) {
      if (re[j] == i) {
        if (reversed) {
          for (long long k = 1; k < j; k++) {
            temp.push_back(re[k]);
            r.push_back(temp);
            temp.clear();
          }
          for (long long k = j; k <= n - i + 1; k++) {
            temp.push_back(re[k]);
          }
          r.push_back(temp);
          temp.clear();
          for (long long k = n - i + 2; k <= n; k++) {
            temp.push_back(re[k]);
            r.push_back(temp);
            temp.clear();
          }
        } else {
          for (long long k = 1; k < i; k++) {
            temp.push_back(re[k]);
            r.push_back(temp);
            temp.clear();
          }
          for (long long k = i; k <= j; k++) {
            temp.push_back(re[k]);
          }
          r.push_back(temp);
          temp.clear();
          for (long long k = j + 1; k <= n; k++) {
            temp.push_back(re[k]);
            r.push_back(temp);
            temp.clear();
          }
        }
        reversed = !reversed;
        break;
      }
    }
    if (r.size() == 1) continue;
    string s = to_string(r.size()) + " ";
    for (auto i : r) {
      s += to_string(i.size()) + " ";
    }
    answers.push_back(s);
    long long a = 1;
    for (long long j = r.size() - 1; j >= 0; j--) {
      for (auto k : r[j]) {
        re[a++] = k;
      }
    }
  }
  if (reversed) {
    string s = to_string(n);
    for (long long i = 0; i < n; i++) {
      s += " 1";
    }
    answers.push_back(s);
  }
  cout << answers.size() << nl;
  for (auto i : answers) {
    cout << i << nl;
  }
}
signed main() {
  long long t = 1;
  while (t--) {
    solve();
  }
}