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

/**
 * 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;
        FastScanner in = new FastScanner(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, FastScanner in, PrintWriter out) {
            int n = in.nextInt();
            int[] a = new int[n];
            for (int i = 0; i < n; i++) {
                a[i] = in.nextInt() - 1;
            }

            boolean stress = false;
//        boolean stress = true;
            if (stress) {
                SplittableRandom random = new SplittableRandom(0);
                n = random.nextInt(1, 53);
                a = new int[n];
                for (int i = 0; i < n; i++) {
                    int j = random.nextInt(i + 1);
                    a[i] = a[j];
                    a[j] = i;
                }
            }

            int[] order = new int[n];
            int p = 0;
            for (int step = 0; step < 2; step++) {
                for (int i = 0; i < n; i++) {
                    if (((i % 2) ^ (n % 2)) != step) {
                        order[p++] = i;
                    }
                }
            }

            int[] who = new int[n];
            for (int i = 0; i < n; i++) {
                who[order[i]] = i;
            }

            int pref = 0;
            int suff = 0;
            List<String> ans = new ArrayList<>();
            for (int step = 0; step < n; step++) {
                int x = -1;
                for (int i = 0; i < n; i++) {
                    if (a[i] == who[step]) {
                        x = i;
                        break;
                    }
                }
                List<Integer> parts = new ArrayList<>();
                if (pref > 0) {
                    parts.add(pref);
                }
                if (x - pref > 0) {
                    parts.add(x - pref);
                }
                parts.add(n - x - suff);
                if (suff > 0) {
                    parts.add(suff);
                }

                if (parts.size() > 1) {
                    apply(a, parts);
                    StringBuilder sb = new StringBuilder();
                    sb.append(parts.size());
                    for (int part : parts) {
                        sb.append(" " + part);
                    }
                    ans.add(sb.toString());
                }

                int npref = suff + 1;
                int nsuff = pref;
                pref = npref;
                suff = nsuff;
            }

            for (int i = 0; i < n; i++) {
                if (a[i] != i) {
                    throw new AssertionError();
                }
            }
            out.println(ans.size());
            for (String s : ans) {
                out.println(s);
            }
        }

        private void apply(int[] a, List<Integer> parts) {
            List<List<Integer>> b = new ArrayList<>();
            int p = 0;
            for (int part : parts) {
                List<Integer> cur = new ArrayList<>();
                for (int i = 0; i < part; i++) {
                    cur.add(a[p++]);
                }
                b.add(cur);
            }
            Collections.reverse(b);
            p = 0;
            for (List<Integer> cur : b) {
                for (int x : cur) {
                    a[p++] = x;
                }
            }
        }

    }

    static class FastScanner {
        private BufferedReader in;
        private StringTokenizer st;

        public FastScanner(InputStream stream) {
            in = new BufferedReader(new InputStreamReader(stream));
        }

        public String next() {
            while (st == null || !st.hasMoreTokens()) {
                try {
                    st = new StringTokenizer(in.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return st.nextToken();
        }

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

    }
}