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:
/*
Written by Kabir Kanha Arora
@kabirkanha
 */

import java.util.*;

public class Main {

    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);

        int n = scanner.nextInt();
        int[] arr = new int[n];
        for (int i = 0; i < n; ++i)
            arr[i] = scanner.nextInt();
        if (checkOrdering(arr, n))
            System.out.println(0);
        else {
            StringBuilder ans = new StringBuilder();
            int countShuffles = 0;
            for (int cnt = 0; cnt < n - 1; ++cnt) {
                if (!checkOrdering(arr, n)) {
                    boolean flag = false;
                    for (int i = 0; i < n - 1 && !flag; ++i) {
                        for (int j = n - 1; j > i && !flag; --j) {
                            if (arr[i] == arr[j] + 1) {
                                for (int k = i; k < j; ++k) {
                                    if (arr[k] > arr[k + 1]) {
                                        ArrayList<Integer> list = new ArrayList<>();
                                        list.add(i);
                                        list.add(k - i + 1);
                                        list.add(j - k);
                                        list.add(n - 1 - j);
                                        int count = 0;
                                        for (int elem : list)
                                            if (elem > 0)
                                                count++;
                                        ans.append(count).append(" ");
                                        for (int elem : list)
                                            if (elem > 0)
                                                ans.append(elem).append(" ");
                                        ans.append("\n");
                                        int[] temp = new int[n];
                                        System.arraycopy(arr, 0, temp, 0, n);
                                        int pos = 0;
                                        for (int x = j + 1; x < n; ++x)
                                            arr[pos++] = temp[x];
                                        for (int x = k + 1; x <= j; ++x)
                                            arr[pos++] = temp[x];
                                        for (int x = i; x <= k; ++x)
                                            arr[pos++] = temp[x];
                                        for (int x = 0; x < i; ++x)
                                            arr[pos++] = temp[x];
                                        countShuffles++;
                                        flag = true;
                                        break;
                                    }
                                }
                            }
                        }
                    }
                }
            }
            System.out.println(countShuffles);
            System.out.println(ans.toString());
        }
    }

    static boolean checkOrdering(int[] arr, int n) {
        for (int i = 0; i < n; ++i)
            if (arr[i] != i + 1)
                return false;
        return true;
    }
}