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]. 
import java.io.*;
import java.util.*;

/**
 * Accomplished using the EduTools plugin by JetBrains https://plugins.jetbrains.com/plugin/10081-edutools
 */

public class Main {
   public static void main(String[] args) throws IOException {
      final BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
      final BufferedWriter bw = new BufferedWriter(new OutputStreamWriter(System.out));

      int n = Integer.parseInt(br.readLine());
      int[] c = new int[n];
      String[] line = br.readLine().split(" ");
      for (int i = 0; i < n; ++i) {
         c[i] = Integer.parseInt(line[i]);
      }
      int[] orig = c.clone();

      int start = 0;
      while (c[start] != 1) {
         ++start;
      }
      List<List<Integer>> operations = new ArrayList<>();
      boolean reversed = false;

      for (int i = 2; i <= n; ++i) {
         int curr = 0;
         while (c[curr] != i) {
            ++curr;
         }
         if (curr == start + i - 1) {
            continue;
         }

         int newC[] = new int[n], newStart;
         boolean flip = reversed;
         List<Integer> d = new ArrayList<>();

         if (curr > start) {
            if (start > 0) {
               d.add(start);
            }
            for (int j = 1; j < i; ++j) {
               d.add(1);
            }
            for (int j = 0; j < start; ++j) {
               newC[start - j - 1] = c[j];
            }
            for (int j = 0; j < i - 1; ++j) {
               newC[start + j] = c[start + j];
            }
            int dif = curr - start - i + 2;
            d.add(dif);
            for (int j = 0; j < dif; ++j) {
               newC[start + i + j - 1] = c[curr - j];
            }
            dif = n - curr - 1;
            if (dif > 0) {
               d.add(dif);
               for (int j = 1; j <= dif; ++j) {
                  newC[n - j] = c[curr + j];
               }
            }
            reversed = !reversed;
            newStart = start;
         } else {
            int index = n;
            if (curr > 0) {
               d.add(curr);
               for (int j = 1; j <= curr; ++j) {
                  newC[--index] = c[curr - j];
               }
            }
            d.add(1);
            newC[--index] = i;
            newStart = index - i + 1;
            int dif = start - curr + i - 2;
            if (dif > 0) {
               d.add(dif);
               for (int j = 0; j < dif; ++j) {
                  newC[--index] = c[curr + dif - j];
               }
            }
            if (index > 0) {
               d.add(index);
               for (int j = n - 1; index > 0; --j) {
                  newC[--index] = c[j];
               }
            }
         }

         if (flip) {
            Collections.reverse(d);
         }

         operations.add(d);
         start = newStart;
         c = newC;
      }
      if (reversed) {
         List<Integer> op = new ArrayList<>();
         for (int i = 0; i < n; ++i) {
            op.add(1);
         }
         operations.add(op);
      }

      bw.write(Integer.toString(operations.size()));
      bw.newLine();
      for (List<Integer> operation: operations) {
         bw.write(Integer.toString(operation.size()));
         for (int i = 0; i < operation.size(); ++i) {
            bw.write(' ');
            bw.write(Integer.toString(operation.get(i)));
         }
         bw.newLine();
      }

      br.close();
      bw.close();
   }
}
