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.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.List;
import java.util.StringTokenizer;



public class TaskD {
    public static void main(String[] arg) {
        final FastScanner in = new FastScanner(System.in);
        final PrintWriter out = new PrintWriter(System.out);
        final int n = in.nextInt();
        final int [] a = in.readIntArr(n);
        Solution solution = new Solution(n, a);
        List<List<Integer>> result = solution.solution();
        out.println(result.size());
        for(List<Integer> l:result) {
            out.println(l.size()+" "+convert(l));
        }
        out.close();
        in.close();
    }

    private static String convert(final List<Integer> a) {
        StringBuilder sb = new StringBuilder();
        for(Integer num:a) {
            sb.append(num).append(' ');
        }
        return sb.toString();
    }

    private static class Solution {
        private final int n;
        private final int [] a;

        public Solution(final int n, final int [] a) {
            this.n=n;
            this.a=a;
        }



        public List<List<Integer>> solution(){
            List<List<Integer>> result = new ArrayList<>();

            while(!inorder(a)) {
                List<int[]> intervals = new ArrayList<>();
                {
                    int i = 0;
                    List<Integer> step = step(a);
                    result.add(step);
                    for (Integer interval : step) {
                        int[] el = new int[interval];
                        for (int j = 0; j < interval; j++) {
                            el[j] = a[i + j];
                        }
                        i += interval;
                        intervals.add(el);
                    }
                }

                {
                    int i = 0;
                    for (int interval = intervals.size() - 1; interval >= 0; interval--) {
                        int[] el = intervals.get(interval);
                        for (int j = 0; j < el.length; j++) {
                            a[i + j] = el[j];
                        }
                        i += el.length;
                    }
                }
            }




            return result;
        }

        private boolean inorder(int[] a) {
            for(int i=1;i<=n;i++){
                if(a[i-1]!=i) return false;
            }
            return true;
        }

        public List<Integer> step(int [] aa){
            List<int []> steps = new ArrayList<>();
            int p = n-1;
            while(p>=0){
                int k = 0;
                for(;k<p;k++){
                    if(a[k]==a[p]+1){
                        break;
                    }
                }
                if(k==p) {

                    while(p>0&&a[p-1]==a[p]-1) {
                        p--;
                    }
                    steps.add(new int[]{p,k});
                    p--;
                    continue;
                }
                int ke = k;
                while(ke<p&&a[ke+1]==a[ke]+1){
                    ke++;
                }
                steps.add(new int[]{ke+1, p});
                steps.add(new int[]{k, ke});
                p=k-1;
            }
            final List<Integer> result = new ArrayList<>();
            for(int i=steps.size()-1;i>=0;i--){
                int [] ft = steps.get(i);
                result.add(ft[1]-ft[0]+1);
            }
            return result;
        }




    }



    private static class FastScanner {
        BufferedReader br;
        StringTokenizer st;

        FastScanner(InputStream stream) {
            try {
                br = new BufferedReader(new InputStreamReader(stream));
            } catch (Exception e) {
                e.printStackTrace();
            }
        }

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

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

        int [] readIntArr(int n){
            int [] result = new int[n];
            for(int i=0;i<n;i++){
                result[i]=Integer.parseInt(next());
            }
            return result;
        }

        void close(){
            try {
                br.close();
            } catch (IOException e) {
                e.printStackTrace();
            }
        }

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

    }

}