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 JAVA):

Solution:
import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.List;
import java.util.Collection;
import java.util.Scanner;
import java.util.Deque;
import java.util.ArrayDeque;
import java.util.Collections;
import java.util.ArrayList;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 */
public class Main
{
    public static void main(String[] args)
    {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        Scanner in = new Scanner(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        TaskD solver = new TaskD();
        solver.solve(1, in, out);
        out.close();
    }

    static class TaskD
    {
        public void solve(int testNumber, Scanner in, PrintWriter out)
        {
            int T = 1;
            while (T-- > 0)
            {
                solveOne(in, out);
            }
        }

        private void solveOne(Scanner in, PrintWriter out)
        {
            int N = in.nextInt();
            int nums[] = CPUtils.readIntArray(N, in);
            int cycles = 0;
            String res = "";
            while (!isSorted(nums))
            {
                cycles++;
                Deque<Integer> split = solve(nums);
                int temp[] = new int[N];

                int idx = nums.length - 1;
                int idxTemp = 0;
                List<Integer> ans = new ArrayList<>();
                while (!split.isEmpty())
                {
                    int len = split.pollLast();
                    if (len == 0) continue;
                    ans.add(len);
                    for (int k = idx - len + 1; k <= idx; k++, idxTemp++) temp[idxTemp] = nums[k];
                    idx -= len;
                }
                nums = temp;
                //System.out.println(Arrays.toString(nums));
                Collections.reverse(ans);
                res = getResult(res, ans);
            }
            out.println(cycles);
            out.println(res);
        }

        private String getResult(String res, List<Integer> ans)
        {
            if (ans.size() > 0)
            {
                res += ans.size() + " ";
                for (int i : ans) res += i + " ";
            }
            res += "\n";
            return res;
        }

        private Deque<Integer> solve(int[] nums)
        {
            Deque<Integer> split = new ArrayDeque<>();
            // case N at the beginning
            if (nums[0] == nums.length)
            {
                split.add(1);
                int i = 0, target = nums.length - 1;
                while (i + 1 < nums.length && nums[i + 1] == nums[i] - 1)
                {
                    i++;
                    split.add(1);
                    target--;
                }
                int cnt = 0;
                for (i++; i < nums.length; i++)
                {
                    cnt++;
                    if (nums[i] == target)
                    {
                        split.add(cnt);
                        cnt = 0;
                    }
                }
                if (cnt > 0) split.add(cnt);
                // case N at the end
            } else if (nums[nums.length - 1] == nums.length)
            {
                split.add(1);
                int i = nums.length - 1, target = nums.length - 1;
                while (i - 1 >= 0 && nums[i - 1] + 1 == nums[i])
                {
                    i--;
                    split.addFirst(1);
                    target--;
                }
                int cnt = 0;
                for (i--; i >= 0; i--)
                {
                    cnt++;
                    if (nums[i] == target)
                    {
                        split.addFirst(cnt);
                        cnt = 0;
                    }
                }
                if (cnt > 0) split.addFirst(cnt);
            } else
            {
                int cnt = 0;
                for (int i = 0; i < nums.length; i++)
                {
                    cnt++;
                    if (nums[i] == nums.length)
                    {
                        split.add(cnt);
                        cnt = 0;
                    }
                }
                if (cnt > 0) split.addLast(cnt);
            }
            return split;
        }

        private boolean isSorted(int nums[])
        {
            for (int i = 0; i < nums.length - 1; i++) if (nums[i] > nums[i + 1]) return false;
            return true;
        }

    }

    static class CPUtils
    {
        public static int[] readIntArray(int size, Scanner in)
        {
            int[] array = new int[size];
            for (int i = 0; i < size; i++)
            {
                array[i] = in.nextInt();
            }
            return array;
        }

    }
}