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.util.*;
import java.io.*;

public class Main {
    public static void main(String[] args) throws IOException {
        FastScanner in = new FastScanner(System.in);
        PrintWriter out = new PrintWriter(System.out);
        new Main().run(in, out);
        out.close();
    }

    public static long mod = 17352642619633L;

    int[] a, posa;
    int[] b, posb;
    void run(FastScanner in, PrintWriter out) {

        int N = in.nextInt();
        a = new int[N];
        posa = new int[N+1];

        for (int i = 0; i < N; i++) {
            a[i] = in.nextInt();
            posa[a[i]] = i;
        }

        b = new int[N];
        posb = new int[N+1];
        List<List<Integer>> ret = new ArrayList<>();

        int k;
        for (int i = 1; i <= N && !sorted(a); i++) {
            LinkedList<Integer> ll = new LinkedList<>();

            k = 0;
            if (i%2 == 0) {
                if (N-(posa[i]+1) != 0) ll.offerFirst(N-(posa[i]+1));
                for (int j = posa[i]+1; j < N; j++) {
                    b[k] = a[j];
                    posb[b[k]] = k;
                    k++;
                }
                if (posa[i]-(i-1)+1 != 0) ll.offerFirst(posa[i]-(i-1)+1);
                for (int j = i-1; j <= posa[i]; j++) {
                    b[k] = a[j];
                    posb[b[k]] = k;
                    k++;
                }
                for (int j = i-2; j >= 0; j--) {
                    ll.offerFirst(1);
                    b[k] = a[j];
                    posb[b[k]] = k;
                    k++;
                }
            } else {

                for (int j = 0; j < i-1; j++) {
                    ll.offerFirst(1);
                    b[k] = a[N-1-j];
                    posb[b[k]] = k;
                    k++;
                }
                if (N-(i-1) - posa[i] != 0) ll.offerFirst(N-(i-1) - posa[i]);
                for (int j = posa[i]; j < N-(i-1); j++) {
                    b[k] = a[j];
                    posb[b[k]] = k;
                    k++;
                }
                if (posa[i] != 0) ll.offerFirst(posa[i]);
                for (int j = 0; j < posa[i]; j++) {
                    b[k] = a[j];
                    posb[b[k]] = k;
                    k++;
                }
            }

            // System.out.println(Arrays.toString(a));
            // System.out.println(Arrays.toString(b));
            // System.out.println(Arrays.toString(ll.toArray()));
            // System.out.println();

            swap();
            if (ll.size() > 1) ret.add(ll);
        }

        out.println(ret.size());
        for (List<Integer> ll : ret) {
            out.print(ll.size() + " ");
            for (int x : ll) out.print(x + " ");
            out.println();
        }



    }

    void swap() {
        int[] tmp = a;
        a = b;
        b = tmp;

        tmp = posa;
        posa = posb;
        posb = tmp;
    }

    boolean sorted(int[] a) {
        for (int i = 0; i < a.length; i++) {
            if (a[i] != i+1) return false;
        }
        return true;
    }

    static class FastScanner {
        BufferedReader br;
        StringTokenizer st;

        public FastScanner(InputStream in) {
            br = new BufferedReader(new InputStreamReader(in));
            st = null;
        }

        String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

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

        long nextLong() {
            return Long.parseLong(next());
        }
    }
}
