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

public class D11 {
    public static void main(String [] args) {
        MyScanner sc = new MyScanner();
        PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));
        int n = sc.nextInt();
        int [] a = new int[n];
        boolean ok = true;
        for (int i = 0; i < n; i++) {
            a[i] = sc.nextInt();
            ok &= (a[i] == i + 1);
        }
        if (ok) {
            out.println(0);
            out.close();
            return;
        }
        StringBuilder sb = new StringBuilder(); int ct = 0;
        while (true) {
            ok = true;
            for (int i = 0; i < n; i++) {
                ok &= (a[i] == i + 1);
            }
            if (ok) break;
            int [] seen = new int[n + 2];
            Arrays.fill(seen, -1); int one = -1; int two = -1;
            for (int i = 0; i < n; i++) {
                if (seen[a[i] + 1] != -1) {
                    two = i;
                    one = seen[a[i] + 1];
                    break;
                }
                seen[a[i]] = i;
            }
            int end = one + 1;
            int b = a[one];
            while (end < n && a[end] == b + (end - one)) {
                ++end;
            }
            int c = 2;
            if (one > 0) c++;
            if (n - two - 1 > 0) c++;
            sb.append(c + " ");
            if (one > 0) sb.append(one + " ");
            sb.append((end - one) + " ");
            sb.append((two - end + 1) + " ");
            if (n - two - 1 > 0) sb.append((n - two - 1) + " ");
            sb.append("\n");
            ArrayList<Integer> before = new ArrayList<>();
            for (int i = 0; i < one; i++) before.add(a[i]);
            ArrayList<Integer> middle = new ArrayList<>();
            for (int i = one; i <= end - 1; i++) middle.add(a[i]);
            ArrayList<Integer> middle2 = new ArrayList<>();
            for (int i = end; i <= two; i++) middle2.add(a[i]);
            ArrayList<Integer> e = new ArrayList<>();
            for (int i = two + 1; i < n; i++) e.add(a[i]);
            int cur = 0;
            for (int i = 0; i < e.size(); i++) a[cur++] = e.get(i);
            for (int i = 0; i < middle2.size(); i++) a[cur++] = middle2.get(i);
            for (int i = 0; i < middle.size(); i++) a[cur++] = middle.get(i);
            for (int i = 0; i < before.size(); i++) a[cur++] = before.get(i);
            ++ct;
        }
        out.println(ct);
        out.println(sb.toString());
        out.close();
    }





    //-----------MyScanner class for faster input----------
    public static class MyScanner {
        BufferedReader br;
        StringTokenizer st;

        public MyScanner() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }

        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());
        }

        double nextDouble() {
            return Double.parseDouble(next());
        }

        String nextLine(){
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }



    }

}