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.StringTokenizer;
import java.io.IOException;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 *
 * @author xwchen
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(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, InputReader in, PrintWriter out) {
            int n = in.nextInt();
            int[] a = in.nextIntArray1(n);
            ArrayList<ArrayList<Integer>> res = new ArrayList<>();
            int cnt = 0;
            while (!isSorted(a) && ++cnt < 100) {
                int x = -1;
                int y = -1;
                for (int i = 1; i <= n && x == -1; ++i) {
                    for (int j = i + 1; j <= n; ++j) {
                        if (a[i] == a[j] + 1) {
                            x = i;
                            y = j;
                            break;
                        }
                    }
                }
                ArrayList<Integer> list = new ArrayList<>();
                if (x > 1) {
                    list.add(x - 1);
                }
                int right = x;
                for (int i = x; i <= n; ++i) {
                    if (i - x == a[i] - a[x]) {
                        right = i;
                    } else {
                        break;
                    }
                }
                list.add(right - x + 1);
                list.add(y - right);
                if (y < n) {
                    list.add(n - y);
                }
                res.add(list);
                deckReverse(a, list);
//            System.out.println(list);
//            System.out.print("cnt = "+cnt+"  ");
//            for (int i = 1; i <= n; ++i) {
//                System.out.print(a[i] + " ");
//            }
//            System.out.println();
            }
            out.println(res.size());
            for (ArrayList<Integer> list : res) {
                out.print(list.size());
                for (int x : list) {
                    out.print(" " + x);
                }
                out.println();
            }
        }

        boolean isSorted(int[] a) {
            for (int i = 1; i < a.length; ++i) {
                if (a[i] != i) return false;
            }
            return true;
        }

        void deckReverse(int[] a, ArrayList<Integer> list) {
            ArrayList<ArrayList<Integer>> collect = new ArrayList<>();
            int j = 0;
            ArrayList<Integer> temp = new ArrayList<>();
            for (int i = 1; i < a.length; ++i) {
                temp.add(a[i]);
                if (temp.size() == list.get(j)) {
                    collect.add(temp);
                    temp = new ArrayList<>();
                    ++j;
                }
            }
            //System.out.println(collect);
            j = 1;
            for (int i = collect.size() - 1; i >= 0; --i) {
                for (int x : collect.get(i)) {
                    a[j++] = x;
                }
            }
        }

    }

    static class InputReader {
        private BufferedReader reader;
        private StringTokenizer tokenizer = new StringTokenizer("");

        public InputReader(InputStream inputStream) {
            this.reader = new BufferedReader(
                    new InputStreamReader(inputStream));
        }

        public String next() {
            while (!tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return tokenizer.nextToken();
        }

        public int nextInt() {
            return Integer.parseInt(next());
        }

        public int[] nextIntArray1(int n) {
            int[] a = new int[n + 1];
            for (int i = 1; i <= n; i++)
                a[i] = nextInt();
            return a;
        }

    }
}