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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.StringTokenizer;

public class Task4 {

    public static void main(String[] args) throws IOException {
        new Task4().solve();
    }

    private void solve() throws IOException {
        BufferedReader f = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter out = new PrintWriter(System.out);

        int n = Integer.parseInt(f.readLine());
        int[] cards = new int[n];
        StringTokenizer tokenizer = new StringTokenizer(f.readLine());
        for (int i = 0; i < n; i++) cards[i] = Integer.parseInt(tokenizer.nextToken());


        ArrayList<ArrayList<Integer>> orders = new ArrayList<ArrayList<Integer>>();
        while (!isOrdered(cards)) {
            //System.out.println(Arrays.toString(cards));
            if (orders.size() > 1000) break;
            int a = -1;
            int b = -1;
            for (int i = 0; i < n; i++) {
                for (int j = 0; j < i; j++) {
                    if (cards[j] == cards[i] + 1) {
                        a = j;
                        b = i;
                    }
                }
            }

            ArrayList<Integer> command = new ArrayList<Integer>();
            /*
            int goodLen = 0;
            for (int i = a - 1; i >= 0; i--) {
                if (cards[i] == cards[i + 1] - 1) goodLen++;
                else break;
            }
            if (a - goodLen != 0) command.add(a - goodLen);
            */
            if (a != 0) command.add(a);
            int goodLen = 1;
            for (int i = a + 1; i < n; i++) {
                if (cards[i] == cards[i - 1] + 1) {
                    goodLen++;
                } else {
                    break;
                }
            }

            command.add(goodLen);
            command.add(b - (a + goodLen - 1));
            if (n - 1 - b != 0) command.add(n - 1 - b);
            cards = performCommand(command, cards);
            orders.add(command);
        }
        //System.out.println(Arrays.toString(cards));

        out.println(orders.size());
        for (ArrayList<Integer> command : orders) {
            out.print(command.size());
            for (int i = 0; i < command.size(); i++) {
                out.print(" " + command.get(i));
            }
            out.println();
        }

        out.close();
    }

    private boolean isOrdered(int[] ar) {
        for (int i = 1; i <= ar.length; i++) {
            if (ar[i -1 ] != i) return false;
        }

        return true;
    }

    private int[] performCommand(ArrayList<Integer> command, int[] cards) {
        int left = 0;
        int right = cards.length - 1;
        int[] newCards = new int[cards.length];
        for (int ar : command) {
            for (int i = left + ar - 1; i >= left; i--) {
                newCards[right] = cards[i];
                right--;
            }
            left += ar;
        }

        return newCards;
    }
}
