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.awt.*;
import java.io.*;
import java.util.*;
import java.util.List;

public class Coding {


    static long MOD = 1000000007;
    static long[] fact = new long[302];
    static long[] invFact = new long[302];
    static long[][][] segmentTree = new long[4*300000 + 11][2][2];
    static long MINVALUE = Integer.MIN_VALUE;


    public static void main(String[] args) {
        BufferedReader bi = new BufferedReader(new InputStreamReader(System.in));
        BufferedWriter writer = new BufferedWriter(new OutputStreamWriter(System.out));
        //precompute();
        try {
            int t = 1;
            while (t > 0) {
                String[] input = bi.readLine().split(" ");
                int n = Integer.parseInt(input[0].trim());
                input = bi.readLine().split(" ");
                int[] ar = new int[n];
                for(int i=0;i<n;++i) {
                    ar[i] = Integer.parseInt(input[i].trim());
                }
                List<List<Integer>> ans = new ArrayList<>();
                while(true) {
                    boolean isSorted = true;
                    for(int i=0;i<n;++i) {
                        if (ar[i]!=(i+1)) {
                            isSorted = false;
                            break;
                        }
                    }
                    if (isSorted) {
                        break;
                    }
                    for(int i=1;i<=n-1;++i) {
                        int first=-1, second=-1;
                        for(int j=0;j<n;++j) {
                            if (ar[j]==i) {
                                first = j;
                            } else if (ar[j]==(i+1)) {
                                second = j;
                            }
                        }
                        if (first>second) {
                            int firstStart = first;
                            int secondLast = second;
                            //System.out.println(firstStart + " " + secondLast);
                            while (true) {
                                if (firstStart == 0) {
                                    break;
                                }
                                if (ar[firstStart-1]!=(ar[firstStart]-1)) {
                                    break;
                                }
                                firstStart--;
                            }
                            while ((secondLast<n-1) && (ar[secondLast+1]==(ar[secondLast]+1))) {
                                secondLast++;
                            }

                            List<Integer> curMove = new ArrayList<>();
                            if (second==0 && first==n-1) {
                                curMove.add(2);
                                curMove.add(secondLast+1);
                                curMove.add(n-secondLast-1);
                                int[] tmp = new int[n];
                                for(int j=0;j<n-secondLast-1;++j) {
                                    tmp[j] = ar[secondLast+1 + j];
                                }
                                int cur = 0;
                                for(int j=n-secondLast-1;j<n;++j) {
                                    tmp[j] = ar[cur];
                                    cur++;
                                }
                                for(int j=0;j<n;++j) {
                                    ar[j] = tmp[j];
                                    //System.out.print(ar[j] + " ");
                                }
                                //System.out.println("In 1");
                            } else if (second==0 && first<n-1) {
                                //System.out.println("Secondlast is " + secondLast);
                                curMove.add(3);
                                curMove.add(secondLast+1);
                                curMove.add(first-secondLast);
                                curMove.add(n-first-1);
                                int[] tmp = new int[n];
                                for(int j=0;j<n-first-1;++j) {
                                    tmp[j] = ar[first + 1 + j];
                                }
                                int cur = 0;
                                for(int j=n-first-1;j<n-secondLast-1;++j) {
                                    tmp[j] = ar[secondLast+1 + cur];
                                    cur++;
                                }
                                cur = 0;
                                for(int j=n-secondLast-1;j<n;++j) {
                                    tmp[j] = ar[cur];
                                    cur++;
                                }
                                for(int j=0;j<n;++j) {
                                    ar[j] = tmp[j];
                                    //System.out.print(ar[j] + " ");
                                }
                                //System.out.println("In 2");
                            } else if (second>0 && first==n-1) {
                                curMove.add(3);
                                curMove.add(second);
                                curMove.add(secondLast-second+1);
                                curMove.add(n-secondLast-1);
                                int[] tmp = new int[n];
                                for(int j=0;j<n-secondLast-1;++j) {
                                    tmp[j] = ar[secondLast + 1 + j];
                                }
                                int cur = 0;
                                for(int j=n-secondLast-1;j<n-second;++j) {
                                    tmp[j] = ar[second+cur];
                                    cur++;
                                }
                                cur=0;
                                for(int j=n-second;j<n;++j) {
                                    tmp[j] = ar[cur];
                                    cur++;
                                }
                                for(int j=0;j<n;++j) {
                                    ar[j] = tmp[j];
                                    //System.out.print(ar[j] + " ");
                                }
                                //System.out.println("In 3");
                            } else {
                                curMove.add(4);
                                curMove.add(second);
                                curMove.add(secondLast-second+1);
                                curMove.add(first-secondLast);
                                curMove.add(n-first-1);
                                int[] tmp = new int[n];
                                for(int j=0;j<n-first-1;++j) {
                                    tmp[j] = ar[first + 1 + j];
                                }
                                int cur = 0;
                                for(int j=n-first-1;j<n-secondLast-1;++j) {
                                    tmp[j] = ar[secondLast + 1 + cur];
                                    cur++;
                                }
                                cur = 0;
                                for(int j=n-secondLast-1;j<n-second;++j) {
                                    tmp[j] = ar[second+cur];
                                    cur++;
                                }

                                cur = 0;
                                for(int j=n-second;j<n;++j) {
                                    tmp[j] = ar[cur];
                                    cur++;
                                }
                                for(int j=0;j<n;++j) {
                                    ar[j] = tmp[j];
                                    //System.out.print(ar[j] + " ");
                                }
                                //System.out.println("In 4");
                            }
                            ans.add(curMove);
                        }
                    }
                }
                writer.append(ans.size() + "\n");
                for(int i=0;i<ans.size();++i) {
                    List<Integer> move = ans.get(i);
                    for(int j=0;j<move.size();++j) {
                        writer.append(move.get(j) + " ");
                    }
                    writer.append("\n");
                }
                writer.flush();
                 t--;
            }
            writer.flush();
        } catch (Exception e) {
            e.printStackTrace();
        }
    }

    private static int gcd(int a, int b) {
        if (a<b) {
            return gcd(b, a);
        }
        if (a%b==0) {
            return b;
        }
        return gcd(b, a%b);
    }

    private static void dfs(int cur, List<Integer>[] adj, int[] nodeCount, int[] maxChild,  int parent, int[] par) {
        par[cur] = parent;
        for(int i=0;i<adj[cur].size();++i) {
            int nxt = adj[cur].get(i);
            if (nxt!=parent) {
                dfs(nxt, adj, nodeCount, maxChild, cur, par);
                nodeCount[cur]+=nodeCount[nxt];
                maxChild[cur] = Math.max(maxChild[cur], nodeCount[nxt]);
            }
        }
    }

    private static int findDist(int a, int b, int[] par) {
        Map<Integer, Integer> dist = new HashMap<>();
        int count = 0;
        dist.put(a, 0);
        while (par[a]!=-1) {
            a = par[a];
            count++;
            dist.put(a, count);
        }
        count=0;
        while(!dist.containsKey(b)) {
            b = par[b];
            count++;
        }
        return count+dist.get(b);
    }

    private static int findLongestPath(int cur, int prev, List<Integer>[] adj, int[] dep) {
        int diameter = 0;
        int highest = -1, second=-1;
        for(int i=0;i<adj[cur].size();++i) {
            int nxt = adj[cur].get(i);
            if (nxt!=prev) {
                diameter = Math.max(diameter, findLongestPath(nxt, cur, adj, dep));
                if (dep[nxt]>highest) {
                    second = highest;
                    highest = dep[nxt];
                } else if (dep[nxt]>second) {
                    second = dep[nxt];
                }
            }
        }
        diameter = Math.max(diameter, highest + second + 2);
        return diameter;
    }

    static long ncr(int n, int r) {
        long num = fact[n];
        long denom = (invFact[r]*invFact[n-r])%MOD;
        //System.out.println("num is " + num + " denom is" + denom);
        return (num*denom)%MOD;
    }

    static long modExp(long a, long b, long mod) {
        //System.out.println("a is " + a + " and b is " + b);
        if (a==1) return 1;
        long ans = 1;
        while (b!=0) {
            if (b%2==1) {
                ans = (ans*a)%mod;
            }
            a = (a*a)%mod;
            b/=2;
        }
        return ans;
    }

    private static void precompute() {
        fact[0] = 1;
        invFact[0] = 1;
        for(int i=1;i<=300000;++i) {
            fact[i] = (fact[i-1]*i)%MOD;
            invFact[i] = modExp(fact[i], MOD-2, MOD);
        }
    }
}